


JOHN ALEXANDER JAMESON, Jr. 

1903-1934 




fcfc4l?*-* ■ * *W tJMsMft 



a!0 



This book belonged to John Alexander Jameson, Jr., A.B., Wil- 
liams, 1925; B.S., Massachusetts Institute of Technology, 1928; 
M.S., California, 1933. He was a member of Phi Beta Kappa, Tau 
Beta Pi, the American Society of Civil Engineers, and the Sigma 
Phi Fraternity. His untimely death cut short a promising career. 
He was engaged, as Research Assistant in Mechanical Engineering, 
upon the design and construction of the U. S. Tidal Model Labora- 
tory of the University of California. 

His genial nature and unostentatious effectiveness were founded 
on integrity, loyalty, and devotion. These qualities, recognized by 
everyone, make his life a continuing beneficence. Memory of him 
will not fail among those who knew him. 




WENTWORTH-SMITH MATHEMATICAL SERIES 



SOLID GEOMETRY 



BY 



GEORGE WEXTWORTir 

AND 

DAVID EUGENE SMITH 



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< gs; s mit h aa > 




r; 

R. 

V 



JM 



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GINN AND COMPANY 

BOSTON • HEW FOBS • 'III' \«.'» • LONDOH 
ATLANTA • DALLAS • COLUMBUS • BAM FRANCISCO 



, \: 

COPYRIGHT, 1888, 1899, BY G. A. WENTWORTH libra** 

COPYRIGHT, 1911, 1913, BY GEORGE WENTWORTH 

AND DAVID EUGENE SMITH 

ENTERED AT STATIONERS' HALL 

ALL RIGHTS RESERVED 

919.1 



ENglNEERlM LIBRAHI 






gfle atftenaum jgregj 

GINN AND COMPANY • PRO- 
PRIETORS • BOSTON • U.S.A. 



PREFACE 

Long after the death of Robert Kecorde, England's first 
great writer of textbooks, the preface of a new edition of 
one of his works contained the appreciative statement that 
the book was " entail' d upon the People, ratified and sign'd 
by the approbation of Time." The language of this sentiment 
sounds quaint, but the noble tribute is as impressive to-day 
as when first put in print two hundred and fifty years ago. 

With equal truth these words may be applied to the Geom- 
etry written by George A. Wentworth. For a generation it 
has been the leading textbook On the subject in America. It 
set a standard for usability that every subsequent writer upon 
geometry has tried to follow, and the number of pupils who 
have testified to its excellence has run well into the millions. 

In undertaking to prepare a work to take the place of the 
Wentworth Geometry the authors have been guided by certain 
well-defined principles, based upon an extended investigation 
of the needs of the schools and upon a study of all that is 
best in the recent literature of the subject. The effects of 
these principles they feel should be summarized for the pur- 
pose of calling the attention of the wide circle of friends of 
the Wentworth-Smith series to the points of similarity and 
of difference in the two works. 

1. Every effort has been made not only to preserve but to 
improve upon the simplicity of treatment, the clearness of ex- 
pression, and the symmetry of page that characterized the 
successive editions of the W.-nt worth Geometry. It 1ms been 
the purpose to prepare a book that should do even more than 
maintain the traditions this work has fostered, 

9754 



iv SOLID GEOMETRY 

2. The proofs have been given substantially in full, to the 
end that the pupil may always have before him a model for 
his independent treatment of the exercises. 

3. To meet a general demand, the number of propositions 
has been decreased so as to include only the great basal theo- 
rems and problems. A little of the less important material 
has been placed in the Appendix, to be used or not as cir- 
cumstances demand. 

4. The exercises, in some respects the most important part 
of a course in geometry, have been rendered more dignified in 
appearance and have been improved in content. The number 
of simple exercises has been greatly increased, while the diffi- 
cult puzzle is much less in evidence than in most American 
textbooks. The exercises are systematically grouped, appear- 
ing in general in full pages, in large type, and at frequent 
intervals. They are not all intended for one class, but are so 
numerous as to allow the teacher to make selections from 
year to year. 

5. The work throughout has been made as concrete as is 
reasonable. Definitions have been postponed until they are 
actually needed, only well-recognized terms have been em- 
ployed, the pupil is led to apply his geometry to practical 
cases in mensuration, and correlation is made with the algebra 
already studied. 

6. All the references to Plane Geometry that are directly 
made in the proof of Solid Geometry have been prefixed to this 
edition so as to be easily accessible. 

The authors are indebted to many friends of the Wentworth- 
Smith series for assistance and encouragement in the labor of 
preparing this edition, and they will welcome any further sug- 
gestions for improvement from any of their readers. 

GEORGE WENTWORTH 
DAVID EUGENE SMITH 



COXTKXTS 



REFERENCES TO PLANE GEOMETRY . 

BOOK VI. LINES AND PLANES IN SPACE 

Links and Pl \m> . 

Dihedral Am. les ..... 

Polyhedral Am. i 

Exercis] ...... 

BOOK VII. POLYHEDRONS, CYLj 5, ANlTCONI 

Polyhedrons . 

Prisms .... 

Parallelepipeds 

Pyramids . . v . 

Regular Polyhedrons . 

Cylinders 

Cones .... 

Exercises 

BOOK VIII. THE SPHERE 






Spheres .... 
Plane Sections and Tangent 

, SPHERK \ l. POL! GONS 

Measurement of Spherical Surfai i - 

MEASl REMENT OF SPHERK \ I. Sol [DS 

Exercises 
APPENDIX 

POL! II I DRONS ..... 

SPHERK \ I SEG mini- 

K'l ,i RE LTIONS «'i ( , BOM BTRY 

History of < Ieome i ry . 



Paoi 

\ii 

.,-■■ 

273 
293 

308 
314 



317 



317 

317 
322 
337 
350 
353 

376 

» / I 
O 1 

L 
•1. 

: 
i 

41 

4- 
o 



INDEX 






SYMBOLS AND ABBREVIATIONS 



= equals, equal, equal to, 

is equal to, or 

is equivalent to. 
> is greater than. 
< is less than. 

II parallel. 
_L perpendicular. 
A angle. 
A triangle. 
O parallelogram. 
□ rectangle. 
O circle, 
st. straight. 
rt. right. 
V since. 
.'. therefore. 



Adj. adjacent. 

Alt. alternate. 

Ax. axiom. 

Const, construction. 

Cor. corollary. 

Def. definition. 

Ex. exercise. 

Ext. exterior. 

Eig. figure. 

Hyp. hypothesis. 

Idem identity. 

Int. interior. 

Post. postulate. 

Prob. problem. 

Prop. proposition. 

Sup. supplementary. 



These symbols take the plural form when necessary, as in the case of 
lis, A, A, CD. 

The symbols +, — , x , ■*- are used as in algebra. 

There is no generally accepted symbol for " is congruent to," and the 
words are used in this book. Some teachers use = or =, and some use 
= , but the sign of equality is more commonly employed, the context 
telling whether equality, equivalence, or congruence is to be understood. 

q.e.d. is an abbreviation that has long been used in geometry for 
the Latin words quod erat demonstrandum, "which was to be proved." 

q. e. f. stands for quod erat faciendum, "which was to be done." 



VI 



REFERENCES TO PLANK GEOMETRY 

28. A portion of a plane bounded l»y tlnvi« straight lines is 
called a triangle. 

41. The whole angular space in a plane about a point is 
• •ailed a perigon. 

52. The following are the most important axioms used in 
geometry : 

1. If equals are added to equals, the sums are equal. 

2. If equals are subtracted from equals, the remainders are 
equal. 

3. If equals are multiplied by equals, the products are equal. 

4. If equals are divided by equals, the quotients are equal. 
In division the divisor is never zero. 

5. Like powers and like positive roots of equals are equal. 

6. If unequals are operated on by positive equals in the 
same way, the results are unequal in the same order. 

7. If unequals are added to unequals in the same order, the 
sums are unequal in the same order ; if unequals are subtracted 
from equals, the remainders are unequal in the reverse order. 

8. Quantities that are equal to the same quantity or to equal 
quantities are equal to each other. 

9. A quantity may be substituted for its equal in an equa- 
tion or in an inequality. 

10. If the first of three quantities is greater than the second, 
and the second is greater than the third, then the first is great, r 
than the third. 

11. The whole is greater than any of its parts, and is equal 
to the sum of all its parts. 

vii 



viii SOLID GEOMETRY 

53. Postulate 5. Any figure may be moved from one place 
1 ) another without altering its size or shape. 

56. All right angles are equal. 

57. From a given point in a given line only one perpendic- 
ular can be drawn to the line. 

60. If two lines intersect, the vertical angles are equal. 

66. Definition of congruent figures. 

67. Corresponding parts of congruent figures are equal. 

88. Two triangles are congruent, if two sides and the included 
angle of the one are equal respectively to two sides and the 
included angle of the other. 

69. Two right triangles are congruent, if the sides of the 
right angles are equal respectively. 

72. Two triangles are congruent, if two angles and the in- 
cluded side of the one are equal respectively to • • • . 

80. Two triangles are congruent, if the three sides of the 
one are equal respectively to the three sides of the other. 

82. Only one perpendicular can be drawn to a given line 
from a given external point. 

84. Of two lines drawn from a point in a perpendicular to a 
given line, cutting off on the given line unequal segments from 
the foot of the perpendicular, the more remote is the greater. 

89. Two right triangles are congruent, if tile hypotenuse and 
a side of ths one are equal respectively to the hypotenuse 
and a side of the other. 

93. Lines that lie in the same plane and cannot meet how- 
ever far produced are called parallel lines. 

94. Through a given point only one line can be drawn par- 
allel to a given line. 

95. Two lines in the same plane perpendicular to the same 
line are parallel. 



REFERENCES TO PLANE GEOMETRY ix 

97. If a line is perpendicular to one of two parallel lines, it 
is perpendicular to the other also. 

112. The sum of any two sides of a triangle is greater than 
the third side, and the difference between any two sales is 
less than the third side. 

116. If two triangles have two sides of the one equal respec- 
tively to two sides of the other, but the third side of the ii: 
triangle greater than the third side of the second, then the 
angle opposite the third side of the first is greater than the 
angle opposite the third side of the second. 

118. A quadrilateral may be a trapezoid, having two Bides 
parallel ; a parallelogram, having the opposite sides parallel ; 
or it may have no sides parallel. 

125. The opposite sides of a parallelogram are equal. 

126. A diagonal divides a parallelogram into two congruent 
triangles. 

127. Segments of parallel lines cut off by parallel lines are 
equal. 

130. If two sides of a quadrilateral are equal and parallel, 

then the other two sides are equal and parallel, and the figure 
is a parallelogram. 

131. The diagonals of a parallelogram bisect each other. 

132. Two parallelograms are congruent, if two sides and the 
included angle of the one are equal respectively to t v. 

and the included angle of the other. 

133. Two rectangles having equal bases and equal altitudes 
are congruent. 

136. The line which joins the mid-points of two sides of a 
triangle is parallel to the third side, and is equal to half the 

third side. 



x SOLID GEOMETRY 

142. Two polygons are 

mutually equiangular, if the angles of the one are equal to 
the angles of the other respectively, taken in the same order ; 

mutually equilateral, if the sides of the one are equal to the 
sides of the other respectively, taken in the same order ; 

congruent, if mutually equiangular and mutually equilateral, 
since they then can be made to coincide. 

145. Each angle of a regular polygon of n sides is equal to 

2(n-2) . 

— * '- right angles. 

n 

146. The sum of the exterior angles of a polygon, made by 
producing each of its sides in succession, is equal to four right 
angles. 

148. To prove that a certain line or group of lines is the 
locus of a point that fulfills a given condition, it is necessary 
and sufficient to prove two things : 

1. That any point in the supposed locus satisfies the con- 
dition. 

2. That any point outside the supposed locus does not satisfy 
the given condition. 

150. The locus of a point equidistant from the extremities 
of a given line is the perpendicular bisector of that line. 

151. Two points each equidistant from the extremities of a 
line determine the perpendicular bisector of the line. 

152. The locus of a point equidistant from two given inter- 
secting lines is a pair of lines bisecting the angles formed by 
those lines. 

159. A closed curve lying in a plane, and such that all of 
its points are equally distant from a fixed point in the plane, 
is called a circle. 



REFERENCES TO PLANE GEOMETRY xi 

162. All radii of the same circle or of equal circles are equal; 
and all circles of equal radii are equal. 

167. In the same circle or in equal circles equal arcs subtend 
equal central angles : and of two unequal arcs the greater sul> 
tends the greater central angle. 

172. In the same circle or in equal circles, if two chords are 
equal, they subtend equal arcs ; and if two chords are unequal, 
the greater subtends the greater arc. 

174. A line through the center of a circle perpendicular to 
a chord bisects the chord and the arcs subtended by it. 

178. In the same circle or in equal circles equal chords are 
equidistant from the center, and chords equidistant from the 
center are equal. 

185. A tangent to a circle is perpendicular to the radius 
drawn to the point of contact. 

195. If two circles intersect, the line of centers is the per- 
pendicular bisector of their common chord. 

204. When a variable approaches a constant in such a way 
that the difference between the two may become and remain 
less than any assigned positive quantity, however small, the 
constant is called the limit of the variable. 

207. If, while approaching their respective limits, two vari- 
ables are always equal, their limits are equal. 

212. In the same circle or in equal circles two central angles 
have the same ratio as their intercepted arcs. 

213. A central angle is measured by the intercepted arc. 

261. In any proportion the product of the extremes is equal 
to the product of the means. 

262. The mean proportional between two quantities is equal 
to the square root of their product. 



xii SOLID GEOMETRY 

269. In a series of equal ratios, the sum of the antecedents 
is to the sum of the consequents as any antecedent is to its 
consequent. 

270. Like powers of the terms of a proportion are in pro- 
portion. 

273. If a line is drawn through two sides of a triangle par- 
allel to the third side, it divides the two sides proportionally. 

274. One side of a triangle is to either of its segments cut 
off by a line parallel to the base as the third side is to its 
corresponding segment. 

275. Three or more parallel lines cut off proportional in- 
tercepts on any two transversals. 

282. Polygons that have their corresponding angles equal, 
and their corresponding sides proportional, are called similar 
polygons. 

285. Two mutually equiangular triangles are similar. 

288. If two triangles have an angle of the one equal to an 
angle of the other, and the including sides proportional, they 
are similar. 

289. If two triangles have their sides respectively propor- 
tional, they are similar. 

290. Two triangles which have their sides respectively par- 
allel, or respectively perpendicular, are similar. 

292. If two polygons are similar, they can be separated 
into the same number of triangles, similar each to each, and 
similarly placed. 

298. If a perpendicular is drawn from any point, on a circle 
to a diameter, the chord from that point to either extremity of 
the diameter is the mean proportional between the diameter 
and the segment adjacent to that chord. 



REFERENCES TO PLANE GEOMETRY xiii 

322. The area of a parallelogram is equal to the product of 

its base by its altitude. 

323. Parallelograms having equal bases and equal altitudes 
are equivalent. 

325. The area of a triangle is equal to half the product of 

its base by its altitude. 

326. Triangles having equal bases and equal altitudes are 
equivalent. 

327. Triangles having equal bases are to each other as their 
altitudes ; triangles having equal altitudes are to each othei 
their bases; any two triangles are to each other as the prod- 
ucts of their bases by their altitudes. 

329. The area of a trapezoid is equal to half the product of 
the sum of its bases by its altitude. 

332. The areas of two triangles that have an angle of the 
one equal to an angle of the other are to each other as the 
products of the sides including the equal angles. 

334. The areas of two similar polygons are to each other as 
the squares on any two corresponding sides. 

377. If the number of sides of a regular inscribed polygon 
is indefinitely increased, the apothem of the polygon approaches 
the radius of the circle as its limit. 

381. The circle is the limit which the perimeters of regular 
inscribed polygons and of similar circumscribed polygons ap- 
proach, if the number of sides of the polygons is indefinitely 
increased. 

The area of the circle is the limit which the areas of the 
inscribed and circumscribed polygons approach. 

382. Two circumferences have the same ratio as their radii 
385. The circumference of a circle equal- 2 irr. 

389. The area of a circle = wr*. 



SOLID GEOMETRY 



BOOK VI 

LINES AND PLANKS IN SPACE 

» • » 

421. The Nature of Solid Gecmetry. la plane geometry we 
deal with figures lying in a §at suiiact , BtjiMj^irjg their proper- 
ties and relations and measuring div figu2< -. In solid geometry 
we shall deal with figures not only of two dimensions but of 
three dimensions, also studying their properties and relations 
and measuring the figures. 

422. Plane. A surface such that a straight line joining any 

two of its points lies wholly in the surface is called a plane. 

A plane is understood to be indefinite in extent, but it is conveniently 
represented by a rectangle seen obliquely, as here shown. 




423. Determining a Plane. A plane is said to be determined 
by certain lines or points if it contains the given lines or 
points, and no other plane can contain them. 

When we suppose a plane to be drawn to include given points or lines, 
we are said to pass the plane through these points or lines. 

When a straight line is drawn from an external point to a plane, its 
point of contact with the plane is called its foot. 

424. Intersection of Planes. The line that contains all the 
points common to two planes is called their intersection. 

273 



274 BOOK VI. SOLID GEOMETRY 

425. Postulate of Planes. Corresponding to the postulate that 
one straight line, and only one, can be drawn through two given 
points, the following postulate is assumed for planes : 

One plane, and only one, can be passed through two given 

intersecting straight lines. 

For it is apparent from the first figure that a plane may be made to 
turn about any single straight line AB, thus assuming different positions. 
But if CD intersects AB at P, as in the second figure, then when the 
plane through AB turns until it includes C, it must include D, since it 
includes two points, C and P, of the line (§ 422) . If it turns any more, it 
will no longer contain C. 





426. Corollary 1. A straight line and a point not in the 
line determine a plane. 

For example, line AB and point C in the above figure. 

427. Corollary 2. Three points not in a straight line deter- 
mine a plane. 

For by joining any one of them with the other two we have two inter- 
secting lines (§ 425). 

428. Corollary 3. Two parallel lines determine a plane. 

M 

C T D A 

A B \ 



N 

For two parallel lines lie in a plane (§ 93), and a plane containing 
either parallel and a point P in the other is determined (§ 426). 



LINES AND PLANES 



27") 



Proposition I. Theorem 

429. If two planes cut each other, their niter section is 

a straight line. 

p N 




Given MN and PQ, two planes which cut each other. 

To prove that the plane* MN and PQ intersect in a 
straight line. 

Proof. Let .4 and B be two points common to the two planes. 
Draw a straight line through the points A and B. 
Then the straight line AB lies in both planes. § 422 
(For it has two points in each plane.) 

No point not in the line AB can be in both planes ; for one 
plane, and only one, can contain a straight line and a point 
without the line. § 426 

Therefore the straight line through A and B contains all 
the points common to the two planes, and is consequently the 
intersection of the planes, by § 424. q.e.d. 

Discussion. What is the corresponding statement in plane geometry ? 

430. Perpendicular to a Plane. If a straight line drawn to a 
plane is perpendicular to every straight line that passes through 
its foot and lies in the plane, it is said to be perpendicular to 

the plane. 

When a line is perpendicular to euplane, the plane is also said to be 
perpendicular to the line. 



276 



BOOK VI. SOLID GEOMETRY 



Proposition II. Theorem 



431. If a line is perpendicular to each of tivo other 
lines at their point of intersection, it is perpendicular 
to the plane of the two lines. 




Given the line AO perpendicular to the lines OP and OR at 0. 

To prove that AO is 1. to the plane MN of these lines. 

Proof. Through draw in MN any other line OQ, and draw 
PR cutting OP, OQ, Oil, at P, Q, and R. 

Produce AO to A', making OA' equal to OA, and join A and 
A ' to each of the points P, Q, and R. 

Then OP and OR are each _L to A A' at its mid-point. 

.'. AP = A'P, and AR = A'R. § 150 

.'. A APR is congruent to A A' PR. § 80 

,'.ZRPA=Z.A'PR. §67 

That is, Z QPA =ZA 'PQ. 



.'. APQA is congruent to APQA'. 



§68 



.-. AQ = A'Q(§ 67); and OQ is _L to AA' at 0. § 151 
.*. AO is _L to any and hence to every line in MN through O. 
.'. AO is -L to the plane MN, by § 430. Q.e.d. 



LINKS AND PLANES 



277 



Proposition III. Theorem 

432. All the perpendiculars that can be drawn to a 
given line at << given point Ha in a plane which is per 
pendictdar to the given Urn at the given point 




R 



1° 



N 



Given the plane MAT perpendicular to the line OY at O. 

To prove that OP, any line ±to OY at 0, lies in MN. 

Proof. Let the plane containing OY and OP intersect the 
plane MN in the line OP'; then OY is ± to OP'. § 430 

In the plane POY only one _L can be drawn to OY at 0. § 57 
Therefore OP and OP' coincide, and OP lies in MN. 
Hence every _L to OY at 0, as OQ, OR, lies in MN. Q.e.d. 

433. I orollary 1. Through a given point in a given line 
one plane, and only one, can be passed perpendicular to the Urn , 

434. Corollary 2. Through agiven externalpoint oneplaru . 
and only one, eon be passed perpendicular to a given line. 

Given the line OY an. 1 the point P. 

Draw PO ± to OY. and OQ ± to OY. 

Then OQ and OP determine a plane through H 1 

PI to OY. / Q L "-->-~P \ 

Only one such plane can be drawn: for N 

only one _L can be drawn to OY from the point P (5 v 

435. Oblique Line. A line that meets a plane but is nut per- 
pendicular to it is said to be oblique to the plane. 



278 



BOOK VI. SOLID GEOMETRY 



Proposition IV. Theorem 

436. Through a given point in a plane there can he 
drawn one line perpendicular to the plane, and only one. 




Given the point P in the plane MN. 

To prove that there can be drawn one line perpendicular to 
tlie plane MN at P, and only one. 

Proof. Through the point P draw in the plane MN any line 
AB, and pass through P a plane IF 1 to AB, cutting the 
plane MN in CD. § 433 

At P erect in the plane XY the line PQ J_ to CD. 

The line AB, being _L to the plane XY by construction, is _L 
to PQ, which passes through its foot in the plane. § 430 

That is, PQ is _L to AB ; and as it is _L to CD by construc- 
tion, it is _L to the plane MN. § 431 

Moreover, any other line PR drawn from P is oblique to 
MN. For PQ and PR intersecting in P determine a plane. 

To avoid drawing another plane, use XY again to represent 
the plane of PQ and PR, letting it cut MN in the line CD. 

Then since PQ is _L to MN, it is _L to Cfi. § 430 

Therefore PR is oblique to CD. § 57 

Therefore PR is oblique to MN. § 435 

Therefore PQ is the only _L to MN at the point P. q.e.d. 

Discussion. What is the corresponding proposition in plane geometry? 



LINES AXD PLANES 



279 



Proposition V. Theorem 

437. Through a given external point there can he draion 
one line perpendicular to a given plane, and only one. 




Given the plane MN and the external point P. 

To prove that there can be drawn one line from P perpen- 
dicular to the plane MX, and only one. 

Proof. In MN draw any line Eft, and let A'}' be a plan*' 
through P _L to Elf, cutting MN in AB, and EH in C. 

Draw PO _L to AB, and in MN draw any line 01) from to Elf. 

Produce PO, making OP' ~0P, and draw PC, PD } P'C, P'D. 

Since DC is _L to X Y, A PCD and P'CD are right angles. § 430 

Since the side DC is common, and PC =P'C, § 150 

.-. rt. A PCD is congruent to rt. A P'CD. 

.'. PD = P'D. 

.-. Of) is_Lto PP'at 0. 

.'.PO is ± to J/.V, being _L to OD and J /;. 

Moreover, every other line PF from P to .l/.V is oblique 
to MN. (The proof is left for tin- student.) 

..PO is the only _L from f> to MN. q.e. d. 

438. Corollarv. 77".' perpendicular is the shortest line 
from a point to a plan*'. 

The- length of this J. is called the distance from the point to the plant-. 



§69 

§67 

§151 

§ i.;i 



280 BOOK VI. SOLID GEOMETRY 

Proposition VI. Theorem 

439. Oblique lines drawn from a point to a plane, 
meeting the plane at equal distances from the foot of 
the perpendicular ', are equal ; and of two oblique lines, 
meeting the prt ane at unequal distances from the foot 
of the perpendicular, the more remote is the greater. 




N 

Given the plane MN, the perpendicular line PO, the oblique lines 
PA, PB, PC, the equal distances OB, OC, and the unequal dis- 
tances OA, OC, with OA greater than OC. 

To prove that PB = PC, and PA > PC. 

Proof. In the A OBP and OCP, 

OP = OP, Iden. 

OB = OC, Given 

and Z.BOP = Z.POC. §56 

.'.A OBP is congruent to A OCP. § 69 

.'.PB = PC. §67 

Let A, B, and O lie in the same straight line. 

Then OA>OC. Given 

.-. OA>OB. Ax. 9 

.-.PA>PB. §84 

.'. PA>PC, by Ax. 9. q.e.d. 
Discussion. Compare the corresponding case in plane geometry. 



LINKS AND PLANES 281 

440. Corollary 1. Equal oblique lines drawn from >> point 
to a plane meet the plane at equal distances from the foot of the 
perpendicular ; and of two unequal oblique lines the greatt t 
meets the plane at the greater distance from the foot of the 
perpi ndicular. 

In the figure on page 280, if PB is given equal to PC, then since 
PO = PO, and the angles at are righl angles, what follows with re- 
spect to the & OBP and OCP? with respect to OB and OC ? 

Furthermore, if PA > PC, how does PA compare with PB? 

Then how does OA compare with OB '.' Why? 

Then how does OA compare with OC ? 

441. Corollary 2. The locus of a point equidistant from 
all points on a circle is a line through the center, perpen- 
dicular to the plane of the circle. 

In the figure on page 280, in order to prove that PO is the required 
locus what must be proved for any point on PO (§ 148) ? for any point 
not on PO ? Prove both of these facts. 

442. Corollary 3. The locus of a point equidistant fmm 

the vertices of a triangle is a line through the center of the 

circumscribed circle, perpendicular to the plane of the triangle* 

How does this follow from Corollary 2 ? 

"What locus is the line through the center of the inscribed circle, per- 
pendicular to the plane of the triangle ? 

443. Corollary 4. The locus of a point equidistant from 
two given points is the plane perpendicular to the line joining 
them, at its mid-point. 

For any point C in this plane lies in a JL to AB 
at 0, its mid-point (§ 430). 

Hence how do CA and CB compare (§ 150) ? 

And any point D outside the plane MN cannot lie 
in a ± to AB at O. What may therefore be sai.l as to the distan 
from Dto A and B (§ 160)? 

What is the proposition in plane geotnetry corresponding to Corol- 
lary 4? In what respect do the two proofs differ ? 




282 



BOOK VI. SOLID GEOMETKY 



Proposition VII. Theorem 

444. Two lines perpendicular to the same plane are 
parallel. 







\ — «r- 



-^rF 



-c^ 



c 



\D 







N 



Given the lines AB and CD, perpendicular to the plane MN. 
To prove that AB and CD are parallel. 

Proof. Draw AD and BD, and in MN draw through D 
EF _L to BD, making DE = DF. Draw BE, AE, BF, AF. 

Now prove that A BDE and BDF are congruent (§ 69), that 
AADE and ADF are right angles (§ 80), and that BD, CD, 
and AD lie in the same plane (§ 432). 

But AB also lies in this plane, § 422 

and AB and CD are both _L to BD. § 430 

.*. AB is II to CD, by § 95. Q.e.d. 

445. Corollary 1. If one of two parallel lines is perpen- 



dicular to a plane, the other is also perpendicular 
to the plane. m 

For if through any point of CD a line is drawn ± to / 
MN, how is it related to AB (§ 444) ? Now apply § 94. 

446. Corollary 2. If two lines are parallel 



o 



B 







b 



N 



A C 



M , 

Q±Z1 

N 



to a third line, they are parallel to each other. 
For a plane MN ± to CD is _L to AB and EF (§ 445) 

447. Line and Plane Parallel. If a line and plane cannot 
meet, however far produced, they are said to be parallel. 



LINES AND PLANKS 283 

EXERCISE 74 

1. Why does folding a sheet of paper give a straight edge ? 

2. If equal oblique lines are drawn from a given external 

point to a plane, they make equal angles with lines drawn from 
the points where the oblique lines meet the plane to tin- toot 
of the perpendicular from the given point. 

3. If from the foot of a perpendicular to a plane a line is 
drawn at right angles to any line in the plane, the line drawn 
from its intersection with the line in the plane to any point 
of the perpendicular is perpendicular to the line of the plane. 

4. If two perpendiculars are drawn from a point to a plane 
and to a line in that plane respectively, the line joining the 
feet of the perpendiculars is perpendicular to the given line. 

5. From two vertices of a triangle perpendiculars are let fall 
on the opposite sides. From the intersection of these perpen- 
diculars a perpendicular is drawn to the plane of the triangle. 
Prove that a line drawn to any vertex of the triangle, from 
any point on this perpendicular, is perpendicular to the line 
drawn through that vertex parallel to the opposite side. 

6. Find the point in a plane to which lines may be drawn 

from two given external points on the same side of the plane 

so that their sum shall be the least possible. 

From cue point A suppose a _L AO drawn to the plane and produced 
to .4'. making OA' — OA. Connect A' and the other point B by a line 
cutting the plane at P. Then BPA is the shortest line. 

7. If three equal oblique lines are drawn from an externa] 
point to a plane, the perpendicular from the point to the plane 
meets the plane at the center of the circle circumscribed about 
the triangle having for its vertices the feet of the oblique lines. 

8. State and prove the propositions of plane geometry cor- 
responding to §§ 444, 44o. and 446. Why do not the proofs 
of those propositions apply to these sections ? 



284 BOOK VI. SOLID GEOMETRY 

Proposition VIII. Theorem 

448. If two lines are parallel, every plane containing 
one of the lines, and only one, is parallel to the other line. 




Given the parallel lines AB and CD, and the plane MN contain- 
ing CD but not AB. 

To prove that the plane MN is parallel to AB. 
Proof. AB and CD are in the same plane, AD. § 93 

This plane AD intersects the plane MN in CD. Given 
Now AB lies in the plane AD, however far produced. § 422 

Therefore, if AB meets the plane MN at all, the point of 
meeting must be in the line CD. § 422 

But since AB is II to CD, Given 

.'. AB cannot meet CD. § 93 

.*. AB cannot meet the plane MN. 

.'. MN is II to AB, by § 447. q.e.d. 

449. Corollary 1. Through either of two lines not in the 
same plane one plane, and only one, can he passed parallel to 
the other. 

A B 

For if AB and CD are the lines, and we pass a 

plane through CD and a line CE which is drawn ^f- 

parallel to AB, what can be said of the plane MN 



determined by CD and CE, with respect to the line \q. 

AB? Why can there be only one such plane ? 2£ 



LINKS AND PLANES 285 

450. Corollary 2. Through a given point one plane, and 

only one, can be passed parallel to any tir<> given lines in space. 

Suppose P the given point and AB and CD the 
given lines. If, now, we draw through 1' the line A 

A'B' parallel to AB, and the line CD' parallel to CD, (f/ p 

these lines will determine the plane MX (§ 425), 



Then what may be said of the plane MX with re- / i P/ B 



speet to the lines AB and CD? Why can only one / C /l' l) ' 
plane be so passed through P ? N 

Discussion. Proposition VIII might of course be made more general 
by allowing both of the parallels to lie in the plane MX. That is, If two 
lines are parallel, a plane containing one of the lines cannot intersect the 
other, although the other line might lie in it. 

In the figure of Corollary 2 the Z D'PB' is sometimes spoken of as the 
angle between the nonintersecting lines AB and CD, although this is not 
commonly done in elementary geometry. 

451. Parallel Planes. Two planes which cannot meet, how- 
ever far produced, are said to be parallel. 



EXERCISE 75 

1. What is the locus of a point in a plane equidistant from 
two parallel lines? What is the corresponding locus in space, 
given two parallel planes instead of two parallel lines ? Draw 
the figure, without proof. 

2. Find the locus in a plane of a point at a given distance 
from a given external point. What is the corresponding case 
of plane geometry ? 

3. If a given line is parallel to a given plane, the intersection 
of the plane with aiiy plane passed through the given line i> 
parallel to that line. 

4. If a given line is parallel to a given plane, a line parallel 

to the given line drawn through any point of the plane lies in 
the plane. 



280 



BOOK VI. SOLID GEOMETRY 



Proposition IX. Theorem 



452. Two planes perpendicular to the same line are 
parallel. 




Given the planes MN and PQ perpendicular to the line AB. 

To prove that the planes MN and PQ are parallel. 

Proof. If MN and PQ are not parallel, they must meet. 
If they could meet, we should have two planes from a point 
of their intersection _L to the same straight line. 

But this is impossible. § 434 

.*. MN and PQ are parallel, by § 451. q.e.d. 



EXERCISE 76 

1. What is the locus of a point equidistant from two given 
points A, B, and also equidistant from two other given points 
C, D? 

2. What is the locus of a point at the distance d from a 
given plane P, and at the distance aV from a given plane P' ? 

3. What is the locus of a point at the distance d from a 
given plane P, and equidistant from two given points A, B? 

4. Find a point at the distance d from a given plane P, at 
the distance d' from a given plane P\ and equidistant from 
two given points A, B. Can there be more than one such 
point ? Draw the figure, without proof. 



LINKS AM) PLANES *2ST 

Proposition X. Theorem 

453. The intersections of 'two parallel planes by a third 



plane are parallel lines. 




Q 
s 

Given the parallel planes MN and PQ, cut by the plane RS in 
AB and CD respectively. 

To prove that the intersections AB and CD an- parallel. 
Proof. AB and CD are in the same plane RS. Given 

If .17? and CD meet, the planes .V.Vand PQ must meet, since 
AJJ is always in MN and CD is always in PQ. § 122 

But MN and PQ cannot meet. § 451 

.-.AB is II to CD, by § 93. Q.e.d. 

454. Corollary 1. Parallel lines included between par- 
allel planes are equal. 

In the above figure, suppose AC II to BD. Then the plane of A C and 
111) will intersect MN and PQ in lines that are how related to each 
other? Then what kind of a figure is ACDB? 

455. Corollary 2. Two parallel planes are everywhere 
equidistant from each other. 

Drop perpendiculars from anypoints in MN to PQ. Prove that these 
perpendiculars are parallel and hence (§ 464) thai they are equal, 



288 



BOOK VI. SOLID GEOMETRY 



Proposition XI. Theorem 

456. A line perpendicular to one of two parallel planes 
is perpendicular to the other also. 




15 



Given the line AB perpendicular to the plane MN, and the plane 
PQ parallel to the plane MN. 

To prove that AB is perpendicular to the plane PQ. 

Proof. Pass through AB two planes AE, AF, intersecting 
MN in AC, AD, and intersecting PQ in BE, BF, respectively. 

Then AC is II to BE, and AD is II to BF. § 453 

But AB is ± to AC and AD. § 430 

.'. AB is _L to BE and BF. § 97 

.*. AB is _L to the plane PQ, by § 431. q.e.d. 

457. Corollary 1. Through a given point one plane, and 
only one, can be passed jmrallel to a given plane. 

How is a plane through A, _L to AB, related to PQ ? Now use § 433. 

458. Corollary 2. The locus of a point equidistant from 
two parallel planes is a plane perpendicular to a line which is 
perpendicular to the planes and which bisects the segment cut 
off by them. 

459. Corollary 3. The locus of a point equidistant from 
two parallel lines is a plane perpendicular to a line which is 
perpendicular to the given lines and which bisects the segment 
cut off by them. 



LINKS AND PLANES 289 

Proposition NIL Theorem 

460. If two intersecting lines are <<"■// jtantllrl t<> <i 
plane, the plane of these lines is parallel to that plane. 



Given the intersecting lines AC, AD, each parallel to the plane 
PQ, and let MN be the plane determined by AC and AD. 

To prove that MN is parallel to PQ. 

Proof. Draw AB _L to PQ. 

Pass a plane through AB and AC intersecting PQ in BE. 
and a plane through AB and AD intersecting PQ in BF. 

Then AB is _L to BE and BF. § 430 

P>ut A C and BE lie in the same plane, Const. 

and .1 C cannot meet BE without meeting the plane PQ, which 
is impossible. 

Similarly 





§447 


.'. BE is li to AC. 


§93 


BF is II to AD. 




AB is _L to AC and to AD. 


§ ( .»7 


AB is JL to the plane MN. 


§431 


MN is 11 to PQ. by § 152. 


Q.E. D. 



Discussion. It is evident that this proposition does not depend upon 
the position of A. For example, C and /> might remain where they are 
and A might recede a long distance, AC and J.D becoming more nearly- 
parallel. So long as the lines intersect, and only so long, are we certain 
that the planes are parallel. 



290 BOOK VI. SOLID GEOMETRY 

Proposition XIII. Theorem 

461. If two angles not in the same plane have their 
sides respectively parallel and lying on the same side of 
the straight line joining their vertices, the angles are 
equal, and their planes are parallel. 




Given the angles A and A\ in the planes MN and PQ respec- 
tively, and their corresponding sides parallel and lying on the same 
side of AA' . 

To prove that ZA = ZA', and that MN is II to PQ. 
Proof. Take AD and A'D' equal, also AC and A'C equal. 

Draw DD', CC, CD, CD'. 
Since AD is equal and II to A'D', 

.'. AA' is equal and II to DD'. § 130 
In like manner A A' is equal and II to CC. 

.'. DD' and CC' are equal, Ax. 8 

and DD' and CC' are parallel. § 446 

.'.CD=C'D'. §130 

.'.A ADC is congruent to AA'D'C. § 80 

.-. AA=Z.A'. §67 

But MN is II to each of the lines A'C and A'D'. § 448 

.'. MN is II to PQ, by § 460. Q.e.d. 

Discussion. Why does not the proof of the corresponding proposition 
in plane geometry apply here ? 



LINES AND PLANES 291 

Proposition XIV. The<>i;i:m 

462. If two lines are cut by three parallel planes > their 
correspond ing segments are proportional. 



Given the lines AB and CD, cut by the parallel planes MN, 
PQ, RS, in the points A, E, B, and C, F, D, respectively. 

To prove that AE : EB = CF : FD. 

Proof. . Draw AD cutting the plane PQ in G. 

Pass a plane through AB and AD, intersecting PQ in the 
line EG, and intersecting RS in the line BD. 

Also pass a plane through AD and CD, intersecting PQ in 
the line GF, and intersecting MN in the line A C. 

Thru EG is II to BD, 

and <7Fis II to AC. §453 

.'. .4/;: EB = AG: GD, 
and CF: FD = A G : GD. § 27.°. 

.-. .1 /:: F/?= CF: 77>. by Ax. 8. q.e.d. 

Discussion. This is a generalization of §275. It may be stated still 
more generally, If tvjo lines are cut by any number of parallel planes, (I. 
* "i-responding segments are proportional. In particular, tin- cum- might be 
considered in which -1 B ami <I) intersect between the plan 

Why does not the proof of the corresponding case (§ 275) in plane 
geometry apply here ? 



292 BOOK VI. SOLID GEOMETRY 

EXERCISE 77 

1. Find the locus of a line drawn through a given point, 
parallel to a given plane. 

2. Find the locus of a point in a given plane that is equi- 
distant from two given points not in the plane. • 

3. Find the locus of a point equidistant from three given 
points not in a straight line. 

4. Find the locus of a point equidistant from two given 
parallel planes and also equidistant from two given points. 

5. What is the locus of a point in a plane at a given dis- 
tance from a given line in the plane ? What is the locus of 
a point at a given distance from a given plane ? 

6. The line AB cuts three parallel planes in the points A, 
E, B ; and the line CD cuts these planes in the points C, F, D, 
If A E = 6 in., EB = 8 in., and CD = 12 in., compute CF and ED. 

7. The line AB cuts three parallel planes in the points A, 
E, B ; and the line CD cuts these planes in the points C, E, D. 
HAB = % in., CF=5 in., and CD = 9 in., compute AE and EB. 

8. To draw a perpendicular to a given plane from a given 
point without the plane. 

9. To erect a perpendicular to a given plane at a given 
point in the plane. 

10. It is proved in plane geometry that if three or more 
parallels intercept equal segments on one transversal, they 
intercept equal segments on every transversal. State and prove 
a corresponding proposition in solid geometry. 

11. It is proved in plane geometry that the line joining the 
mid-points of two sides of a triangle is parallel to the third 
side. State and prove a corresponding proposition in solid 
geometry, referring to a plane passing through the mid-points 
of two sides of a triangle. 



DIHEDRAL AM-I.KS 



-'.»:; 



463. Dihedral Angle. The opening between two intersecting 

plant's is called a dihedral angle. 

In this figure the two planes AM 

and BN are called the farts of the 
dihedral angle, and the line of inter- 
section AB is called the edge of the 
anide. 

A dihedral angle is read by nam- 
ing the letters designating its edge, 

or its faces and edge, or by a small letter within. Thus the 
angle here shown may be designated by AB, M-AH-X. or <l. 




dihedral 




464. Size of a Dihedral Angle. The si/..- of a dihedral ai, 
depends upon the amount of turning- necessary to bring one 
face into the position of the other. 

The analogy to the plane angle is apparent, and is still further seen 

as we proceed. 

465. Adjacent Dihedral An- 
gles. If two dihedral angles 
have a common edge, and a 
common face between them, 
they are said to be adjacent 
dihedral angles. 

For example, M-AB-X and X-BA-P are adjacent dihedral angles. 

466. Right Dihedral Angle. If one plane meets another plane 
and makes the adjacent dihedral angles equal, each of these 
angles is called a right dihedral angle. 

Dihedral angles are said to be straight, acute, obtuse, reflex, compU- 
meniary, supplementary, conjugate, &nd vertical, under conditions similar 

to those obtaining with plane angles. There is little occasion, howeyer, 
to nse any of these terms in connection with dihedral ang 

467. Perpendicular Planes. If two planes intersect and form 
a right dihedral angle, each of the planes is said to be perjh 
dicular to the other plan.-. 







B 


X 


A 


B' 


X 


A' 


B" 


V 





294 BOOK VI. SOLID GEOMETRY 

468. Plane Angle of a Dihedral Angle. The plane angle formed 
by two straight lines, one in each plane, perpen- O r 
dicular to the edge at the same point, is called 
the plane angle of the dihedral angle. 

For example, A A OB is the plane angle of the dihedral 
angle 00", if AO and BO are each _L to 00". 

9 

469. Corollary. The plane angle of a dihedral angle has 
the same magnitude from whatever point in the edge the per- 
pendiculars are drawn. 

How is O'B' related to OB, and O'A' to OA (§ 95) ? Then how is 
AA'O'W related to ZAOB (§ 461) ? 

470. Relation of Dihedral Angles to Plane Angles. It is appar- 
ent that the demonstrations of many properties of dihedral 
angles are identically the same as the demonstrations of anal- 
ogous properties of plane angles. A few of the more important 
propositions will be proved, but the following may be assumed 
or may be taken as exercises : 

1. If a plane meets another plane, it forms with it two adjacent 
dihedral angles whose sum is equal to two right dihedral angles. 

2. If the sum of two adjacent dihedral angles is equal to two right 
dihedral angles, their exterior faces are in the same plane. 

3. If two planes intersect each other, their vertical dihedral angles 
are equal. 

4. If a plane intersects two parallel planes, the alternate-interior dihe- 
dral angles are equal ; the exterior-interior dihedral angles are equal ; 
and the two interior dihedral angles on the same side of the transverse 
plane are supplementary. 

5. When two planes are cut by a third plane, if the alternate-interior 
dihedral angles are equal, or the exterior-interior dihedral angles are 
equal, and the edges of the dihedral angles thus formed are parallel, 
the two planes are parallel. 

6. Two dihedral angles whose faces are parallel each to each are 
either equal or supplementary. 

7. Two dihedral angles whose faces are perpendicular each to each, 
and whose edges are parallel, are either equal or supplementary. 



DIHEDRAL ANGLES 



295 



Proposition XV. Theorem 

471. Tivo dihedral angles are equal if their plane 
angles are equal. 

Ex -zsmC' 





Given two equal plane angles ABD and A'B'D' of the two dine* 
dral angles d and d' . 

To prove that the dihedral angles d unci d' are equal 

Proof. Apply dihedral angle d' to dihedral angle d, making 
the plane Z A'B'D' coincide with its equal A ABD. 

Then since B'C is _L to A'B' and D'B', 

.'. B'C is _L to the plane A'B'D'. 

.'. B'C will also be _L to the plane ABD at B. 

.-. B'C will fall on BC. 



§468 

§431 

Post. 5 

§436 



Then the planes A'B'C and ABC, having in common the 
two intersecting lines AB and BC, coincide. § 125 

In the same way it may be shown that the planes D'B'C 
and DBC coincide. 

Therefore the two dihedral angles d and d 1 coincide and are 

equal Q ED 

Discussion. May we have equal straight dihedral angles ? equal reflex 

dihedral angles'.' What is the authority fur saying that right dihedral 
angles are equal ? 



296 



BOOK VI. SOLID GEOMETRY 



Proposition XVI. Theorem 

472. Two dihedral angles have the same ratio as their 
plane angles. 




Fig. 1 Fig. 2 Fig. 3 

Given two dihedral angles 2?C and B'C, and let their plane angles 
be ABD and A'B'D' respectively. 

To prove that A B'C :ABC=.A A'B'D' : A ABD. 

Case 1. When the plane angles are commensurable. ' 

Proof. Suppose the A ABD and A'B'D' (Figs. 1 and 2) have 
a common measure, which is contained m times in A ABD and 
n times in A A'B'D'. 

Then A A'B'D' : A ABD = n: m. 

Apply this measure to A ABD and A A'B'D', and through 
the lines of division and the edges BC and B'C pass planes. 

These planes divide ABC into m parts, and A B'C' into n 
parts, equal each to each. § 471 

.'. Z B'C : ABC = n : m. 

.'. AB'C : ABC = AA'B'D' : A ABD, by Ax. 8. q.e.d. 

As with plane angles, there is also the case of incommensurables. 
Since the common measure may be taken as small as we please, it is 
evident that for practical purposes the above proof is sufficient. The 
proof for the incommensurable case, p. 297, may be omitted at the 
discretion of the teacher without destroying the sequence. 



DIHEDRAL ANGLES 297 

Case 2. When thr />l<tne angles <m- i n < ommensurable. 

Proof. Divide the Z.ABD into any number of equal parts, 
and apply one of these parts to the Z.A'B'D 1 | Figs. 1 and 3) 
as a unit of measure. 

Since A. Mil) and /.A'B'D' are incommensurable, a certain 
number of these parts will form the ZA'B'E, leaving a re- 
mainder Z EB'D', less than one of the parts. 

Pass a plane through B'E and B'C 

Since the plane angles of the dihedral angles A-BC-D and 
A'-B'C'-E are commensurable, 

.*. A'-B'C'-E: A-BC-D = Z A' I'll: A ABD. Case 1 

By increasing the number of equal parts into which Z.ABD 
is divided we can diminish the magnitude of each part, and 
therefore can make the Z EB'D' less than any assigned positive 
value, however small. 

Hence the Z EB'D' approaches zero as a limit, as the number 
of parts is indefinitely increased, and at the same time the cor- 
responding dihedral Z E-B'C'-D' approaches zero as a limit. § 204 

Therefore the A A' B'E approaches the A A'B'D' as a limit, 
and the Z A'-B'C'-E approaches the AA'-B'C'-D' as a limit. 

,i Z.17;7<; , Z . 17*7/ 

.-. the variable , A r>r , approaches as a limit, 

Z.ABD ll AABD 

and the variable . " ' approaches ^ \_i> ( <_d aS a limit " 

„ Z.lTTE . ' A A'-B'C'-E , .,_,__ 

But Z ^^ D is always equal to -^ — — > as Z ,1 S£ 

varies in value and approaches A A'B'D 1 as a limit. Case 1 

AA'-B'C'-D' A A'B'D' nv ^ 

■> by § 20 i. q.e.d. 



Z A-BC-D AABD 



473. Corollary. 77"' p&ww "//.<//'• of a dihedral angle may 
be taken as the measure of the dihedral angle. 



298 



BOOK VI. SOLID GEOMETRY 



Proposition XVII. Theorem 

474. If two planes are perpendicular to each other, a 
line drawn in one of them perpendicular to their inter- 
section is perpendicular to the other. 




Given the planes MN and PQ perpendicular to each other, and the 
line CD in PQ perpendicular to their intersection AB. 

To prove that CD is perpendicular to the plane MN. 

Proof. In the plane MN draw DE _L to AB at D. 

Then Z EDC is a right angle, § 473 

and Z. CD A is also a right angle. Given 

.*. CD is _L to the plane MN, by § 431. q.e.d. 

475. Corollary 1. If two planes are perpendicular to each 

other, a perpendicular to one of them at any point of their 

intersection will lie in the other. 

Will a line CD drawn in the plane PQ ± to AB at D be ± to the plane 
MN? How many Js can he drawn from D to the plane MN? 

476. Corollary 2. If two planes are perpendicular to each 

other, a perpendicular to the first from any point in the second 

will lie in the second. 

Will a line CD drawn in the plane PQ from C _L to AB be ± to the 
plane MN ? How many Js can be drawn from C to the plane MN ? 



DIHEDRAL ANGLES 



299 



Propositi* >\ Will. Theorem 

477. If a Vim is perpendicular to a plane, even/ plane 
passed through this line is perpendicular to tJu plane. 




Given the line CD perpendicular to the plane MN at the point Z), 
and PQ any plane passed through CD intersecting MN in AB. 

To prove that the plane PQ is perpendicular to the plane MN. 

Proof. Draw l)E in the plane JAVJLto AB. 

CD is J_ to MX, 

.'. CD is _L to AB. . 

.'.Z.EDC measures Z X-AB-P. 

But Z EDC is "a right angle. 

.-. PQ is J. to MN, by §467. 



Since 



Given 
§430 
§ 173 
§430 

Q.E.D. 



EXERCISE 78 

1. A plane perpendicular to the edge of a dihedral angle is 
perpendicular to each of its faces. 

2. If one line is perpendicular to another, is any plane passed 
through the first line perpendicular, to the second '.' Prove it. 

3. If three lines are perpendicular to one another at a com- 
mon point, what is the relation. to out* another of the tin ■ 

planes determined by the three pairs of lines? Prove it. 



300 



BOOK VI. SOLID GEOMETRY 



Proposition XIX. Theorem 



478. If tivo intersecting planes are each perpendicular 
to a third plane, their intersection is also perpendicular 
to that plane. 




Given two planes BC and BD, intersecting in AB y and each per- 
pendicular to the plane PQ. 

To prove that AB is perpendicular to the plane PQ. 
Proof. Let the plane BC intersect the plane PQ in BF, 
and let the plane BD intersect the plane PQ in BE. 
From any point A on AB draw AX _L to BE, 

and from A draw A Y _L to BF. 
Then AX and A Y are both J_ to the plane PQ. § 474 
But it is impossible to draw two Js to the plane PQ 

from a point outside the plane PQ, § 437 

or from a point in the plane PQ. § 436 

.*. AX and AY must coincide. 

But AX and A Y can coincide only if they lie in both planes. 

And all points common to both planes lie in AB. § 429 

.*. AX and A Y coincide with AB. 

.'. AB is _L to the plane PQ. q.e.d. 

Discussion. How does it appear from this proof that AB cannot be 
parallel to PQ ? 

The proposition is illustrated in the intersection of two walls of a room 
with the floor or the ceiling. - - 



DIJIKDKAL ANCLKS 



301 



Proposition XX. Theorem 

479. The locus of a point equidistant from tJu faces 

<■/" a il'ilii'ili'itl </><<//< is thi plane bisecting tfu angle. 

A' 




Given the plane AM bisecting the dihedral angle formed by the 
planes AD and AC. 

To prove tlmt tin- plum- AM in tlf focus of a point equi- 
distant from the planes AD and AC. 

Proof. Lei EOF be a plane _L to .1 0, the intersectioD of the 

planes A D and \< \ at 0. 

Since AO is _L to the plane EOF, 
.-. the planes .1 D, AM. and AC are ± to the plane EOF. § 477 

From any point P, in the intersection of tin* plant- AM and 
EOF, draw PF J_ to OF, and PE _L to OF. 

Then PF is _L to AD, and /'/■; is J_ to AC. § 171 

.-. /'/•' and /'/•,' measure the distances from the point P to 
the planes .1 D and AC. § l ;;s 

Since 4 is J_ to OF, OP, and 0A\ 130 

.-. (>P bisects Z /'"/:. § 173 

.*. OP is the locus of a point equidistant from "/'and OE. § L62 

.-. 4ilf, wln.-h contains all point- /•. is the locus of a point 
equidistanl from the planes AD and AC. q.e.d. 



302 



BOOK VI. SOLID GEOMETRY 



Proposition XXI. Theorem 

480. Through a given line not perpendicular to a given 
plane, one plane and only one can he passed perpen- 
dicular to the plane. 




Given the line AB not perpendicular to the plane MN. 

To prove that one plane can be passed through AB perpen- 
dicular to the plane MN, and only one. 

Proof. From any point X of AB draw ATI to the plane MN, 
and through AB and XY pass a plane A P. § 425 

The plane AP is _L to the plane MN, since it passes through 
AT, a line _L to MN. § 477 

Moreover, if two planes could be passed through AB _L to the 
plane MN, their intersection AB would be JL to MN. § 478 

But this is impossible, since AB is not _L to MN Given 

Hence one plane can be passed through AB ± to the plane 
MN, and only one. q.e.d. 

481. Projection of a Point. The foot of the line from a given 
point perpendicular to a plane is called the 
projection of the point on the plane. 

482. Projection of a Line. The locus of the 
projections of the points of a line on a plane 
is called the projection of the line on the plane. 




DIHEDRAL ANGLES 



:\\):\ 



Proposition XXII. Theorem 

483. Hie projection of a straight line riot perpendicu- 
lar to a plane, upon that plane, is a straight Urn . 




§ 125 

§ 177 



Given the straight line AB not perpendicular to the plane MN, 
and A'B' the projection of AB upon MN.- 

To prove that A'B' is a straight line. 

Proof. From any point X of AB draw A'}' _L to MX, 
and pass a plane AP through XY and AB. 
The plane .IP is _L to the plane MX, 
and contains all the Js drawn from AB to MX. § 176 
Hence A'B' must be the intersection of these two planes. 

Therefore A'B' is a straight line, by § 429. q.e.d. 

484. Corollary. The projection of a straight line perpen- 
dicular to a plane, upon that plane, is a point. 

485. Inclination of a Line. The angle which a line makes 
with its projection on a plane is considered as the angle which 
it makes with the plane, and is called the inclination of tin- 
line to the plane. 

Therefore a line ordinarily makes an acute angle with a plane, since 
it makes an acute angle with its projection on the plane. The case- 
perpendicular and parallel lines have already been considered. 



304 BOOK VI. SOLID GEOMETRY 

Proposition XXIII. Theorem 

486. The acute angle which a line makes with its 
projection upon a plane is the least angle which it 
makes with any line of the plane. 



N 

Given the line AB meeting the plane MN at A , AB' being the pro- 
jection of AB upon the plane MN, and AD being any other line drawn 
through A in the plane MN. 

To prove that Z B'AB is less than Z DAB. 

Proof. Make AD equal to AB', and draw BB' and BD. 
Then in A BAB' and BAD, 

AB = AB, Iden. 

AB' = AD, Const. 

and BB'<BD. §438 

.'. Z B'AB < Z DAB, by § 116. Q.e.d. 

Discussion. Since Z B'AB is the least angle that AB makes with any 
line of the plane, how does ABAC compare with the angles that AB 
makes with other lines of the plane ? State the general proposition 
involved in the answer. 

If AB is parallel to the plane, what interpretation may be given to 
the proposition ? 

If AB is perpendicular to the plane, what interpretation -may be given 
to the proposition ? 

As AB swings around from the position AB' to the position A C, what 
kind of change takes place in the angle DAB? 



DIHEDRAL ANGLES 305 

EXERCISE 79 

1. Describe tin- position of a segment of a lint* relative to ;i 

given plant- it the projection of tin- segment on the plane is 
equal to its own length. 

2. From a point I. 4 in. from a plane MX. an oblique line 
AC 5 in. long is drawn to the piano and made to turn around 
the perpendicular A B dropped from .1 to the plane. Find the 
area of the circle described by tin- point C. 

3. From a point . 1 . 8 in. from a plane MX. a perpendicular .1 B 
is drawn to the plant-; with /; as a center ami a radius equal 
t<> 6 in., a circle is described in the plant-: at any point Con 
this circle a tangent CD is drawn 24 in. in length. Find the 
distance from .1 to I>. 

4. Equal lines drawn from a given external point to a given 
plane are equally inclined to the plane. 

5. If three equal lines are drawn to a plane from an exter- 
nal point, the perpendicular from the point to the plant- deter- 
mines the center of the circle circumscribed about the triangle 
determined by the planes of the three lines. 

6. Three lines not in the same plane meet in a point. How 
shall a line be drawn so as to make equal angles with all thi 
of these lines '.' 

7. From a point /'two perpendiculars PX and PFare drawn 
to two planes MN and A C which intersect in All. From }' a 
perpendicular YZ is drawn to MX. Prove that the line XZ is 
perpendicular to AB. 

8. If the length of the shadow of a tree standing on level 
ground exceeds the height ofthe tree, the angle made by the 
sun above the horizon must be less than what known angL 

9. Find the locus of a point at a given distance from a given 
plane and equidistant from two given points not in the plane. 



306 



BOOK VI. SOLID GEOMETRY 



Proposition XXIV. Theorem 

487. Between two lines not in the same plane there 
can he one common perpendicular, and only one. 




Given AB and CD, two lines not in the same plane. 

To prove that there can be one common perpendicular, and 
only one, between AB and CD. 

Proof. Through any point A of AB draw AG II to DC. 

Let MN be the plane determined by AB and AG. § 425 

Then the plane MN is II to DC. § 448 

Through DC pass the plane PQ _L to the plane MN. § 480 

Then DC cannot meet D'C, since it is II to the plane MN and 

lies in the plane PQ. § 422 

.'. DC is II to D'C. § 93 

.-. if AB is II to D'C it must be II to DC. § 446 

But AB is not II to DC, for they are not in the same plane. Given 

.*. AB must intersect D'C at some point as C. 

Draw C'C 1 to the plane MN. 

Then C'C is _L to AB and to D'C. § 430 

Since C'C is _L to D'C, and lies in plane PQ, § 475 

.-.C'C is _L to DC. §97 

Therefore one common perpendicular can be drawn. 
It remains to be proved that no other can be drawn. 



DIHEDRAL ANGLES 307 

If it were possible that another common perpendicular could 
be drawn, we might suppose /•;.! to be _L to both 42? and CD. 

Then EA would be _L to A G } § ( .»7 

and therefore EA would be _L to the plane MX. § 431 

BtslwEE'± to D'C. 
Then EE' is J_ to the plane MN. §474 

But this is impossible, if EA is also J_ to the plane MN. § 437 

Hence the supposition that there is a second common per- 
pendicular, EA, leads to an absurdity. 

Therefore there can be one common perpendicular, and only 
one, between AB and CD. q. e.d. 

488. Corollary. The common perpendicular between two 
lines not in the same plane is the shortest line joining them. 

How does CC compare in length with EE' ? Why ? 
How does EE' compare in length with EA ? 

EXERCISE 80 

1. Parallel lines have parallel projections on a plane. 

2. If two planes are perpendicular to each other, any linp 
perpendicular to one of them is how related to the other '.' 

3. If three lines passing through a given point P are cut by 
a fourth line that does not pass through P, the four lines all 
lie in the same plane. 

4. Seven lines, no three of which lie in the same plane, 
pass through the same point. How many planes are deter- 
mined by these lines ? 

5. A cubical tank 10 in. deep contains water to a depth of 
7 in. A foot rule is placed obliquely <>n the bottom so as just 
to reach the top edge of the tank. Make a sketch of the tank, 
and compute the length of the rule covered by water. 



308 BOOK VI. SOLID GEOMETRY 

489. Polyhedral Angle. The opening of three or more planes 
which meet at a common point is called a polyhedral angle. 

The common point V is called the vertex of the angle ; V 

the intersections VA, VB, etc., of the planes are called yf\ 

the edges; the portions of the planes lying between the // \ 

edges are called the faces; and the angles formed by "4f~y£. \ 
adjacent edges are called the face angles. f- — — -4(7 

Every two adjacent edges form a face angle, and every 
two adjacent faces form a dihedral angle. The face angles and dihedral 
angles are the parts of the polyhedral angle. 

490. Size of a Polyhedral Angle. The size of a polyhedral 
angle depends upon the relative position of its faces, and not 
upon their extent. 

491. Convex and Concave Polyhedral Angles. A polyhedral 
angle is said to be convex or concave according as a section 
made by a plane that cuts all its edges at other points than 
the vertex is a convex or concave polygon. 

Only convex polyhedral angles are considered in this work. 

492. Classes of Polyhedral Angles. A polyhedral angle is called 
a trihedral angle if it has three faces, a tetrahedral angle if it 
has four faces, and so on. 

Other names, like pentahedral, hexahedral, heptahedral, etc., for 
angles with 5, 6, 7, etc., faces, are rarely vised. 

A polyhedral angle is designated by a letter at the vertex, or by let- 
ters representing the vertex and all the faces taken in order. Thus, in 
the above figure the trihedral angle is designated by V or by V-ABC. 
A tetrahedral angle would be designated by V or by V-ABCD. 

493. Equal Polyhedral Angles. If 

the corresponding parts of two poly- 
hedral angles are equal and are ar- 
ranged in the same order, the poly- ' l£ ^Jl- ' Y---~--\n' 
hedral angles are said to be equal. 

Thus the angles V-ABC and V'-A'B'C are equal. Equal polyhedral 
angles may evidently be made to coincide by superposition. 



POLYHEDRAL ANGLES 309 

Proposition XXV. THEOREM 

494. The sum of any two face angles of a trihedral 
angle is greater than the third fain angle* 

v 



'r 

Given the trihedral angle V-XYZ, with the face angle XVZ greater 
than either of the face angles XVY or YVZ. 

To prove that Z XVY+ Z YVZ is greater them Z XVZ. 

Proof. In the Z X VZ draw VW, making Z XV W = Z XVY. 
Through any point D of VW draw ADC in the plane XVZ. 

On VY take VB equal to VD. 

Pass a plane through the line AC and the point /;. 

Then since A V = A V, VD = VB, and Z A VD = Z A VB, 

.'. A A VD is congruent to A A VB. § 68 

.\AD = AB. §67 

In the A. 4 £C, AB + BOAC. §112 

Since \B=AD, .\BC>DC. Ax. 6 

In the A £T'C and DFC, 

PC= TC, and F£ = TO, but BC > DC. 

.-. ZBVCis greater than Z.DVC. §11< ! 

.'. Z^ VB + Z.BVC is greater than Z.I FD + ZD7C. Ax. 6 

But Z.AVD + /.DVC = Z.AVC. Ax. 11 

.'. Z A VB + Z.BVC is greater than Z.I I \ '.» 

That is, Z XVY + Z FKZ is greater than Z \T/ q.e.d. 



310 BOOK VI. SOLID GEOMETRY 

Proposition XXVI. Theorem 

495. The sum of the face angles of any convex poly- 
hedral angle is less than four right angles. 




Given a convex polyhedral angle V, all of its edges being cut by 
a plane making the section ABCDE. 

To prove that AAVB + ABVC, etc., is less than four rt. A. 
Proof. From any point P within the polygon draw PA, PB, 

PC, PD, PE. 

The number of the A having the common vertex P is the 
same as the number, having the common vertex V. 

Therefore the sum of the A of all the A having the common 
vertex V is equal to the sum of the A of all the A having the 
common vertex P. 

But in the trihedral A formed at A, B, C, etc., 

Z EA V + Z.BAV is greater than Z BAE, 

Z.VBA + Z CBV is greater than Z CBA , etc. § 494 

Hence the sum of the A at the bases of the A whose com- 
mon vertex is V is greater than the sum of the A at the bases 
of the A whose common vertex is P. Ax. 7 

Therefore the sum of the A at the vertex V is less than the 
sum of the A at the vertex P. Ax. 7 

But the sum of the A at P is equal to 4 rt. A. § 41 

Therefore the sum of the A at V is less than 4 rt. A. q.e.d. 



POLYHEDRAL ANGLES 



311 




496. Symmetric Polyhedral Angles. It" the faces of a poly- 
hedral angle V-ABCD arc produced through the vertex l\ 
another polyhedral angle V-A'B'C'D* is formed, symmetric with 

respect to /- V-A Bt I >. 

The face angles .I VB, BVC, 

etc., are equal respectively to 
the face angles A'VB', /M'< . 
etc. (§ 00). 

Also the dihedral angles VA, 
VB, etc., are equal respectively 
to the dihedral angles VA', VB', 
etc. (§470). (The second figure 
shows a pair of these vertical 
dihedral angles.) 

Looked at from the point V. the edges of Z. V-A1UD are arranged 
from left to right (counterclockwise) in the order VA, VB, \'<\ PD,bat 
the edges of AV-A'B'C'T)' are arranged from right to left (clockwi 
in the order VA', VB', VC, VD'; that is. in an order the reverse of the 
order of the edges in Z V-ABCD. Therefore, 

Two symmetric polyhedral angles have all their parts equal each to each 
hut arranged in reverse order. 

497. Symmetric Polyhedral Angles not Superposable. In gen- 
eral, two symmetric polyhedral angles are not superposable. 
Thus, if the trihedral angle V-A'B'C is made to 
turn 180° about AT, the bisector of the angle 
CVA', then VA 1 will coincide with VC, VC" with 
VA, and the face A'VC" with AVC ; but the di- 
hedral angle VA, and hence the dihedral angle 
VA', not being equal to VC, the plane A'VB' will 
not coincide with BVC] and. for a similar reason, 
the plane C'VB' will not coincide with -1 VB. Hence the ed 
\'B' takes some position VB" not coincident with VB\ that is, 
the trihedral angles are not superposable. 

An analogous case is seen in a pair of gloves. All the pai one 

are equal to the corresponding parts of the other, but the right-hand 
glove will not fit the left hand. 




312 BOOK VI SOLID GEOMETRY 

Proposition XXVII. Theorem 

498. Tioo trihedral angles are equal or symmetric 
when the three face angles of the one are equal respec- 
tively to the three face angles of the other. 



B' E' A' B E A A' E'B' 

Given the trihedral angles V and V\ the angles BVA } CVA, CVB 
being equal respectively to the angles B'V'A', C'V'A', CVB'. 

To prove that the angles V and V are equal or symmetric. 

Proof. On the edges of these angles take the six equal seg- 
ments VA, VB, VC, V'A', V'B', V'C. 

Draw AB, BC, CA, A'B', B'C, C'A'. 

The isosceles ABA V, CA V, CBV are congruent respectively 
to the isosceles AB'A'V, C'A'V, C'B'V. § 68 

.'. AB, BC, CA are equal respectively to A'B', B'C, C'A'. § 67 

.'.ABAC is congruent to AB'A'C. § 80 

From any point D in VA draw DE in the face A VB and DF 
in the face AVC, each _L to VA. 

These lines meet AB and AC respectively. 

(For the AVAB and VAC are acute, each being one of the equal 

A of an isosceles A.) 

Draw EF. 
On A'V take A'D' equal to AD. 



POLYHEDRAL ANGLES 313 

Draw D'E' in the face . 1 ' V'B' and D'F 1 in the face .1 ' \''C, each 
_L to I'M', and draw E'F\ 

Then since AJ> = A'l>', Const. 

and Z />.!/•; = Z I>'.\ 'E', §67 

.'. rt. A .4/)£ is congruent to rt, A I 'D'E'. § 71' 

.-. .1 B = A'E', and 7>/^ = ///•;'. § 67 

In like manner A E=A'E', and DF= D'j". 

Furthermore, since it has been proved that 

ABAC is congruent to AB'A'C, 

.'. £CAB = Z.C'A'B'. § 67 

.'. AA EE is congruent to A A'F'E'. § 68 



.\EF=E'E'. §67 

.'.A EDE is congruent to AE'D'F'. § 80 

.-. ZEDE = ZE'D'E'. § 67 

.". dihedral Z F.4 = dihedral Z I'M '. § 47:; 

(For A FDE and FITE', the measures of these dihedral A, are equal.) 

In like manner it may be proved that the dihedral angles 
VB and VC are equal respectively to the dihedral angles V'B 1 
and V'C. 

.*. the trihedral angles V and V' are equal, § 493 

or else they are symmetric, by § 496. o.p.D. 

This demonstration applies to either of the two figures denoted by 
Y'-A'B'C which are symmetric -with respect to each other. If the first 
of these figures is taken, V and V are equal. If the second is taken, 
V and V are symmetric. 

499. Corollary. If two trihedral angles }></>-,' the tin-,, 

/</<-,■ ,1/,,/Irs <,f tl,e one eqn<il respectively to th>> three face 

angles of the other, thin the dihedral angles of the one are 

equal respectively to th>- dihedral angles of the other. 

For whether the trihedral angles are equal or symmetric, - • ir»-<l in 
the proposition, the dihedral angles are equal (§§ 403, 496). 



314 BOOK VI. SOLID GEOMETRY 

EXERCISE 81 

1. Find the locus of a point in a space of three dimensions 
equidistant from two given intersecting lines. 

2. Find a point at equal distances from four points not all 
in the same plane. 

3. Two dihedral angles which have their edges parallel and 
their faces perpendicular are equal or supplementary. 

4. The projections on a plane of equal and parallel line- 
segments are equal and parallel. 

5. Two trihedral angles are equal when two dihedral angles 
and the included face angle of- the one are equal respectively 
to two dihedral angles and the included face angle of the other, 
and are similarly placed. 

6. Two trihedral angles are equal when two face angles and 
the included dihedral angle of the one are equal respectively 
to two face angles and the included dihedral angle of the other, 
and are similarly placed. 

7. If the face angle A VB of the trihedral angle V-ABC is 
bisected by the line VD, the angle CVD is less than, equal to, 
or greater than half the sum of the angles AVC and BVC, 
according as Z. C VD is less than, equal to, or greater than 90°. 

* 8. If two face angles of a trihedral angle are equal, the 
dihedral angles opposite them are equal. 

9. A trihedral angle having two of its face angles equal is 
superposable on its symmetric trihedral angle. 

10. Find the locus of a point equidistant from the three edges 
of a trihedral angle. 

11. Find the locus of a point equidistant from the three faces 
of a trihedral angle. 

12. The planes that bisect the dihedral angles of a trihedral 
angle meet in a straight line. 



EXERCISES 315 

EXERCISE 82 

Problems of Computation 

1. From a point P, 4 in. from a plane, a line PX is drawn 
meeting tlic plane at X. If PX is 5 in., what is the length of 
the locus of X in the plane '.' 

2. From a point /'. 5 in. from a plane, a line PX is drawn 
meeting the plane at X. It'/'.v is 1- in., what area is inclosed 
in the plane by the locus of X? Answer to two decimal plac< 

3. The base AB of the isosceles triangle ABC in the plane 

MX is 6 in., and the perimeter of the triangle is L'O in. If the 
triangle revolves about its base as an axis, what is the great* 
distance from the plane that is reached by ( ' '.' Answer to I h] 
decimal places. 

4. Two points A and B are 4 in. apart. A point P in<> 

as to be constantly 5 in. from each of these points. Find the 
length of the locus of P. Answer to three decimal plac< 

5. Two parallel planes MN and PQ are cut by a third plane 
RS so as to make one of the dihedral angles 27° 15' 30". Find 
the other dihedral angles. 

6. Two lines are cat by three parallel planes. The segments 
cut from one line are 3 in. and 5^ in., and these cut from the 
other line are 7^ in. and .'•. Find the value of .>■. 

7. Two given planes I angles to each other. A 
point A' is 8 in. from each plan . How far is .V from the e.i 
of the right dihedral igle ' 

8. What is the lengtl I the projection on a plane of a line 
whose length is 10 ~v ~. the inclination of tne line to the plane 
being I.V ? 

9. From the external point P a perpendicular /'/''. 9 in. long. 
is drawn to a plane M V. From P the line PQ is drawn to the 
plane making the angle P'PQ equal to 30°. Find the length of 
the projection of PQ on the plane M V. 



316 BOOK VI. SOLID GEOMETRY 

EXERCISE 83 
Review Questions 

1. How many and what conditions determine a straight line ? 
How many and what conditions determine a plane ? 

2. What simple numerical test, following the measurement 
of certain lengths, determines whether or not one line is perpen- 
dicular to another ? a line is perpendicular to a plane ? 

3. How many planes can be passed through a given line 
perpendicular to a given plane ? Is this true for all positions 
of the given line ? 

4. Through a given point how many lines can be drawn 
parallel to a given line ? parallel to a given plane ? Through 
a given point how many planes can be passed parallel to a 
given line ? parallel to a given plane ? 

5. What is the locus, in a line, of a point equidistant from 
two given points ? in a plane ? in a space of three dimensions ? 

6. What is the locus, in a plane, of a point equidistant from 
two intersecting lines ? State a corresponding proposition for 
solid geometry. 

7. What may be said of two lines in one plane perpendicular 
to the same line ? State two corresponding propositions for 
solid geometry. Does one of these propositions state that two 
planes perpendicular to the same plane are parallel ? 

8. What may be said of a line perpendicular to one of two 
parallel lines ? State two corresponding propositions for solid 
geometry. Is a plane perpendicular to one of two parallel 
planes perpendicular to the other ? 

9. If a line is perpendicular to a plane, what may be said 
of every plane passed through this line ? Does a true prop- 
osition result from changing the word "perpendicular" to 
" parallel" in this statement? 



BOOK VII 

POLYHEDRON *. CYLINDERS, WD CONES 

500. Polyhedron. A solid bounded by planes Is called & poly- 
hedron. 

For example, the figures on pages :i 1 7 and 318 are polyhedrons. 

The bounding planes are called the faces of the polyhedron, the in- 
tersections of the faces are called the edges <»f tin- polyhedron, and the 
intersections of the edges are called the vertices of the polyhedron. 

A line joining any two vertices not in the same fair is called a 
diiHjonal of the polyhedron. 

The plural of polyhedron \& polyhedrons or polyhedra. 

501. Section of a Polyhedron. It a plane passes through a 

polyhedron, the intersection of the plane with such laces as it 
cuts is called a section of the polyhedron. 

502. Convex Polyhedron. If every section oi a polyhedron 
is a convex polygon, tin- polyhedron is said t<> be convex. 

Only convex polyhedrons are considered in this work. 

503. Prism. A polyhedron of which two faces are congruent 

polygons in parallel planes, tin- other lac, 
being parallelograms, is called a prism. 

The parallel polygons are called the bases of the 
prism, the parallelograms arc called the lateral 
faces, and the intersections of the lateral fac< 
are called the lateral edges. 

The sum of the areas of the lateral faces is 
called the lateral area of the prism. 

The lateral edges of a prism are equal (£ 125). 

504. Altitude of a Prism. The perpendicular distance 

tween the planes of the l»;i>es of a prism is called its altitude. 

::i7 




318 



BOOK VII. SOLID GEOMETRY 



505. Right Prism. A prism whose lateral edges are per- 
pendicular to its bases is called a right 
prism. 

The lateral edges of a right prism are equal to 
the altitude (§455). 

506. Oblique Prism. A prism whose lat- 
eral edges are oblique to its bases is called 
an oblique prism. 

507. Prisms classified as to Bases. Prisms 
are said to be triangular, quadrangular, 
and so on, according as their bases are 
triangles, quadrilaterals, and so on. 




Right Prism 




508. Right Section. A section of a prism 
made by a plane cutting all the lateral edges / 
and perpendicular to them is called a right 

'section. Oblique Triangular Prism 

In the case of oblique prisms it is sometimes necessary to produce 
some of the edges in order that the cutting plane may intersect them. 





Truncated Prism 



Eight Section of a Prism 

509. Truncated Prism. The part of a prism included between 
the base and a section made by a plane oblique to the base is 
called a truncated prism. 



PRISMS 319 

Proposition I. Theorem 

510. Tin sections of a prism modi by parallel planes 
cutting all the lateral edges are congruent polygon 




Given the prism PR and the parallel sections AD, A'D' cutting 
all the lateral edges. 

To prove that AI) is congruent to A'D'. 

Proof. -AB is II to A'B', BC is II to B'( '. ( h is to C'D\ 

and so on for all the corresponding sidi § 153 

.-. .1/; = A'B', BC = B'(", CD= CD', 
and so on for all the corresponding sid ? 1-7 

and Z.CBA = ZC'B'A' i Z.DCB = Z D'C'B\ 

and so on for all the corresponding angL 161 

.-. AD is congruent to A'D\ by § 1 l_ q.e. d. 

Discussion. I- the proof the same whether or not the two parallel 
planes are parallel to the basi 

If the sections are all parallel to the bases, are they also congruent to 
the bases ? 

Would the proposition be true if the prism were concave instead of 
convex ? 

Suppose the bases were squares, what would be known as to the form 
of the sectioi 

511. Corollary. Every section of a prism made by a 
plane parallel to the bas* is congruent to th* ■•■ .' and all 
right sections of a prism an wngruent 



320 



BOOK VII. SOLID GEOMETRY 



Proposition II. Theorem 

512. Hie .lateral area of a prism is equal to the 
product of a lateral edcje bij the perimeter of a rigid 
section. 





Given VWXYZ a right section of the prism AD', I the lateral 
area, e a lateral edge, and p the perimeter of the right section. 

To prove that I = ep. 

Proof. A A' = BB' = CC' = DD' = EE' = e. § 503 

Furthermore, VW is _L to BB', WX to CC', XY to DD', YZ 

to EE', and ZV to A A'. § 508 

.'. the area of O AB' = BB' X VW = e X VW y § 322 

the area of OBC'=CC'x WX = e X WX, 

the area of O CD' = DD' x XY = e X XY, and so on. 

But I is equal to the sum of these parallelograms. § 503 

.-. I == e (VW + WX + XY+YZ + ZV). 
But VW + WX + X Y+YZ + ZV =p. 

.-. 1= ep, by Ax. 9. 

513. Corollary. The lateral area of a right prism is 

equal to the product of the altitude by the perimeter of the base. 

Tor how would p then compare with AB + BC + CD + DE + EA ? 
The truth of the corollary is easily seen hy imagining the right prism 
laid on one of its lateral faces, and the surface as it were unrolled. 



Ax. 1 
Ax. 11 

Q.E.D. 



PRISMS 323 

EXERCISE 84 

Find the lateral areas of the right prisms whose altitudes 
and i" rirm U rs of basi s an as follows : 

1. a=18in.,jp =29 in. 4. a = 1 ft. 7 in., p = 2 ft. 9 in. 

2. a = 22 in., p = .">7 in. 5. a = •"» ft. 8 in../' = 5 ft. 7 in. 

3. a = 4.25 in../> = 6.75 in. 6. a =12 ft. '1 in../, == L'7 ft. 9 in. 

Find the lateral areas of the prisms whose lateral edges and 
perimeters of right sections are as follows: 

7. e =17 in., j? = 27 in. 10. e=l ft. 3 in., p = 2 ft. 3 in. 

8. i =23 in.,^? = 35 in. 11. e=2 ft. 7 in., p = 3 ft. ( .» in. 

9. e = 2| in.,// = 4; in. 12. e= 6ft. 1\ in., p = 8ft. 9 J in. 

JV^/7 the lateral edges of the prisms whose lat ral areas and 
perimeters of right sections are as follows: 

13. £=187 sq. in.,]) =11 in. 

14. £ = 357 sq. in., p = 21 in. 

15. 2=169 sq. in..y/=l ft. 1 in. 

16. The lateral surface of an iron bar 5 ft. long is to be 
gilded. The right sectioo is a square whose area is 2.89 sq. in. 
How many square inches of gilding are required '.' 

17. A right prism of glass is 2\ in. long. Its righl ion 
is an equilateral triangle whose altitude is 0.866 in. i .1 \ 3 in. 
Find the lateral surface. 

18. Find the total area of a right prism whose base -pun.' 
with area 5.29 sq. in., and whose length is twice its thicknefi 

19. What is the total area of a right prism whose altitude 
is 32 in., and whose base is a righl triangle with hypotenuse 
106 in. and with one side 84.8 in.? 

20. Every section of a prism mad.- by a plane parallel t.» the 
lateral edges is a parallelogram. 



322 



BOOK VII. SOLID GEOMETRY 



514. Parallelepiped. A prism whose bases are parallelograms 
is called a parallelepiped. 

The word is also, with less authority, spelled parallelopiped.' 

515. Right Parallelepiped. A parallelepiped whose edges are 
perpendicular to the bases is called a right parallelepiped. 

516. Rectangular Parallelepiped. A right parallelepiped whose 
bases are rectangles is called a rectangular parallelepiped. 

By §§ 430 and 453 the f our lateral faces are also rectangles. 




Rectangular Parallelepiped Cube Oblique Parallelepiped 

517. Cube. A parallelepiped whose six faces are all squares 
is called a cube. 

We might also say that a hexahedron whose six faces are all squares 
is a cube, because such a figure would necessarily be a parallelepiped. 

518. Unit of Volume. In measuring volumes, a cube whose 

edges are all equal to the unit of length is taken as the unit 

of volume. 

Thus, if we are measuring the contents of a box of which the dimen- 
sions are given in feet, we take 1 cubic foot as the unit of volume. If the 
dimensions are given in inches, we take 1 cubic inch as the unit. 

519. Volume. The number of units of volume contained by 
a solid is called its volume. 

520. Equivalent Solids. If tAvo solids have equal volumes, 
they are said to be equivalent. 

521. Congruent Solids. If two geometric solids are equal in 
all their parts, and their parts are similarly arranged, the solids 
are said to be congruent. 



PARALLELEPIPEDS 



323 



Proposition III. Theorem 

522. Two prisms an congrw nt if the thn e faa s which 
include << trihedral angle of the one an respectively con- 
gruent to three faces which includi a trihedral angh oj 
the other, and are similarly placed. 




Given the prisms AI and AT, with the faces AD, AG, ^/re- 
spectively congruent to A'D', A'G', A'J', and similarly placed. 

To prove thai AI is congruent to .17'. 

Proof. The face ABAE, BAF, EAF are equal to the face 
. B'A'E', B'A'F', E'A'F' respectively. § 1 12 

Therefore the trihedral angles .1 and .1' are equal. 198 

Apply tin- trihedral angle .1 to its equal A'. 

Then the face AD coincides with A'D\ AG with A'G 1 , and 
A.i with .!'./'; and C falls al C", and D at D'. 

The lateral edges of the prisms are parallel. § I L6 

Therefore CH falls along C'lV, and Dl along D'l'. § 94 

Since the points /', G, and J coincide with /■''. G\ and J\ 
each to each, the planes of the upper bases coincide. § 4l'7 

Hence // coincides with //', and / with /'. 

Hence the prisms coincide and are congruent, hy § 521. q. e. d. 

523. Corollary 1. Two truncated prisms are congruent 
under the conditions giv< n in Proposition III. 

524. Corollary 2. Two right prisms having congruent 

/"/.sfs and rijual altitudes an congruent 



824 BOOK VII. SOLID GEOMETRY 

Proposition IV. Theorem 

525. An oblique prism is equivalent to a right prism 
whose base is equal to a right section of the oblique 
prism, and whose altitude is equal to a lateral edge of 
the oblique prism. 




Given a right section FI of the oblique prism AD\ and FI' a 
right prism whose lateral edges are equal to the lateral edges of AD'. 

To prove that AD' is equivalent to FI 1 . 

Proof. If from the equal lateral edges of AD' and FI' we 
take the lateral edges of FD', which are common to both, the 
remainders AF and A'F', BG and B'G', etc., are equal. Ax. 2 
The bases FI and F'l' are congruent. § 510 

Place A I on A 'I' so that FI shall coincide with F'l'. 
Then FA, GB, etc., coincide with F'A', G'B', etc. § 436 

Hence the faces GA and G'A ', IIB and H'B', coincide. 
But the faces FI and F'l' coincide. 
.'. the truncated prisms A I and A' I' are congruent. § 523 
.'. AI + FD' = A'l' + FD'. Ax. 1 

But . AI+FD' = AD', 

and A 'I' + FD' == FI 1 . Ax. 11 

Therefore AD' is equivalent to FI', by Ax. 9. q.e.d. 



PARALLELEPIPEDS 325 

Proposition V. Theorem 



526. The opposite fares of a parallelepiped are con- 
gruent and paralUI. 



B 
Given a parallelepiped ABCD-A'B'C'D' . 

To prove that the opposite faces AB' and DC are con- 
gruent and parallel. 

Proof. AB is II to DC, § 118 

and AB=DC. §125 

Likewise J . I ' is II and equal to DD\ 

,\ZBAA' = ZCDD'. § 461 

.-. AB' is II to DC'. § 161 

.-. AB' is congruent to DC', by § 132. q.e.d. 

EXERCISE 85 

1. If in the above figure the three plane angles at A are 
80°, 70°, 75°, what are all the other angles in the faces ? 

2. Given a parallelepiped with the three plane angles at 
one of the vertices 85°, 75°, 60°, to find all the other angles 
in the faces. 

3. Given a rectangular parallelepiped lettered as in the fig- 
ure above, and with AB= 4, BC = 3, and CC' = 3j, to find the 
length of the diagonal AC 1 . 

4. The four diagonals of a rectangular parallelepiped are 
equal 

5. Compute the lengths of the diagonals of a rectangular 
parallelepiped whose edges from any vertex are a, b, a. 



326 



BOOK VII. SOLID GEOMETRY 



Proposition VI. Theorem 

527. Hie plane passed through two diagonally oppo- 
site edges of a parallelepiped divides the parallelepiped 
into tivo equivalent triangular prisms. 





Given the plane ACCA 1 passed through the opposite edges A A' 
and CC of the parallelepiped AC. 

To prove that the parallelepiped AC is divided into tivo 
equivalent triangular prisms ABC-B' and AC D-D 1 . 

Proof. Let WXYZ be a right section of the parallelepiped. 
The opposite faces AB' and DC' are parallel and equal. § 526 
Similarly, the faces AD 1 and BC' are parallel and equal. 

.'. WX is II to ZY, and WZ to XY. § 453 

Therefore WXYZ is a parallelogram. § 118 

The plane ACC'A' cuts this parallelogram WXYZ in the 
diagonal WY. § 429 

.'.A WXY is congruent to A YZW. § 126 

Mow shall it be proved that prism ABC-B' is equivalent to 
a right prism with base WXY and altitude .4.4' ? 

How shall it be proved that prism CDA-D' is equivalent to 
a right prism with base YZW and altitude A A' ? 

How are these two right prisms known to be equivalent ? 

How does this prove the proposition ? 

Discussion. What is the corresponding proposition of plane geometry ? 



PARALLELEPIPEDS 327 

EXERCISE 86 

1. The lateral luces of a right prism are rectangles. 

2. The diagonals of a parallelepiped bisect one another. 

3. The three edges of the trihedral angle at one of the ver- 
tices <>f a rectangular parallelepiped are 5 in.. 6 in., and 7 in. 
respectively. Required the total area of the six faces of the 
parallelepiped. 

4. The three lace angles at one vertex of a parallelepiped 
an- each 60°, and the three edges of the trihedral angle with 
that vertex are 3 in., 2 in., 1 in. respectively. Required the 
total area of the six faces. Answer to two decimal places. 

5. In a rectangular parallelepiped the square on any diag- 
onal is equivalent to the sum of the squares on any three edges 
that meet at one of the vertices. 

6. In a box 3 in. deep and 6 in. wide a wire 1 ft. long can 
be stretched to reach from one corner to the diagonally oppo- 
site corner. Required the length of the box. Answer to two 
decimal places. 

7. The diagonal of the base of a rectangular parallelepiped 
is 31f in. and the height of the parallelepiped is 23.7 in. 
Required the length of the diagonal of the parallelepiped. 

8. The total area of the six faces of a cube is 18 sq. in. 
Find the diagonal of the cube. 

9. The diagonal of the face of a cube equals Vl4. Find 
the diagonal of the cube. 

10. The diagonal of a rube equals 2.75 V§. Find the diagonal 
of a face of the cube. 

11. .V water lank is 3 ft. long, 2 ft. 6 in. wide, and 1 ft. 9 in. 
deep. How many square feet of zinc will be required to line 
the four sides and the base, allowing 1? sq. ft. for overlapping 

and for turning t lie top edge '.' 



328 



BOOK VII. SOLID GEOMETRY 
Proposition VII. Theorem 



528. Tivo rectangular parallelepipeds having con- 
gruent bases are to each other as their altitudes. 




Given two rectangular parallelepipeds P andP', with congruent 
bases and with altitudes AB and A'B'. 

To prove that P : P'= AB : A'B'. 

Case 1. When AB and A'B' are commensurable. 

Proof. Suppose a common measure of AB and A'B' to be 
contained m times in AB, and n times in A'B'. 

Then AB : A'B' = m : n: 

Apply this measure to AB and A'B', and through the several 
points of division pass planes perpendicular to these lines. 

These planes divide the parallelepiped P into m parallele- 
pipeds and the parallelepiped P' into n parallelepipeds, con- 
gruent each to each. § 524 

.'. P :P' — m'.n. 

.-. P:P' = AB: A'B', by Ax. 8. Q.E.D. 

The proof for the incommensurable case is similar to that in other 
propositions of this nature. It may be omitted at the discretion of the 
teacher without destroying the sequence, if the incommensurable cases 
are not being considered by the class. 



PARALLELEPIPEDS 

Cask '2. When AB and A'B' are incommensurable, 

r 



:\'1V 




Proof. Divide AB into any number of equal parts, and apply 

one of these parts to J'/;' as a unit of measure as many tin, 
as A'B' will contain it. 

Sine.' AB and A'B' are incommensurable, a certain number 
of these parts will extend from A' to a point D, leaving a 
remainder hi'.' less than <>n.- of the parts. 

Through I> pass a plane X to. 17;', and lei Q denote the 
parallelepiped whose base is the same as that of J'', ami who 
altitude is .1 'D. 



Then 



Q:P=:A'D: AB. 



< lase t 



If the number of parts into which M> is divided is indefi 
nitely increased, the ratioQ:P approaches P':P as a limit, 
and the ratio A 'D: AB approaches A'B': M> as a limit § 204 

The remainder of the proof of the incommensurable case 
is substantially as in the proof given on page297, and it is 
therefore left for the student. 

529. Dimensions. The lengths of the three edges of a rec- 
tangular parallelepiped which meet at a common vertex are 
called its dimensions. 

530. Corollary. Two rectangular parallelepipeds which 
have two dim % in common an to each other as their third 
dimensions. 



330 



BOOK VII. SOLID GEOMETRY 



Proposition VIII. Theorem 

531. Two rectangular parallelepipeds having equal 
altitudes are to each other as their hases. 




Given two rectangular parallelepipeds, P and P f , and a, &, c, and 
a\ b\ c, their three dimensions respectively. 

P ab 



To prove that 



P'" a'V 



Proof. Let Q be a third rectangular parallelepiped whose 
dimensions are a', b, and c. 

Now Q has the two dimensions b and c in common with P, 
and the two dimensions a' and c in common with P\ 



Therefore 
and 



P <L_ 

Q~"a~'' 
P' = Z b 1 ' 



§530 



The products of the corresponding members of these two 
equations give p i 

-JTi = 1777 ' ^ Ax - ' 3 - 



a'b' 



Q.E.D. 



532. Corollary. Two rectangular parallelepipeds which 
have one dimension in common are to each other as the 
products of their other two dimensions. 

For any edge of a rectangular parallelepiped may be taken as the 
altitude, whence § 531 applies. 



PARALLELEPIPEDS 



331 



Proposition IX. Theorem 

533. Two rectangular parallelepipeds an to each other 
as tfo products of their thm dimensions. 




Given two rectangular parallelepipeds, P and P\ and a, b, c, and 
a', b\ d , their three dimensions respectively. 

/ ' abc 
P' "~ a'b V ' 



To prove that 



Proof. Lei Q be a third rectangular parallelepiped whi 
dimensions are ". b\ and c. 

ac 

P' 



Then 

and 



a'c' 



§ 530 



, • - • > . > 






Q.E.D. 



534. Corollary 1. Th volurrn of a rectangular parallel 

piped is equal to th product of its thra dimensions. 

For in the abovi b'= r' = 1. thm /" = 1 x 1 • l- 1 (§ 518). 

But the volume of P(§ 519) ia P : /' . and P P abc : 1 (§ There- 

fore the volume of P is a&c. 

535. Corollary 2. '/'//.■ /•„/// m« of a rectangular parallele- 
piped is equal to the product and altitude. 

r r the volume of P is oi . and aft equals the base and c the altitude. 



332 



BOOK VII. SOLID GEOMETRY 



Proposition X. Theorem 

536. The volume of any parallelepiped is equal to the 
product of its base by its altitude. 




Given an oblique parallelepiped P of volume v, with no two of 
its faces perpendicular, with base b and with altitude a. 

To prove that v = ba. 

Proof. Produce the edge EF and the edges II to EF, and cut 
them perpendicularly by two parallel planes whose distance 
apart GI is equal to EF. We then have the oblique parallele- 
piped Q whose base c is a rectangle. 

Produce the edge IK and the edges II to IK, and cut them 
perpendicularly by two planes whose distance apart MN is 
equal to IK. We then have the rectangular parallelepiped R. 

Now P = Q, and Q = R. § 525 

.\P = R. Ax. 8 

The three parallelepipeds have a common altitude a. § 455 

Also b = c, § 323 

and c = d. § 133 

.'.b = d. Ax. 8 

But the volume of it = da. § 535 

Putting P for R, and b for d, we have v = ba, by Ax. 9. Q. e. d. 

537. Corollary. The volume of any parallelepiped is equal 
to that of a rectangular parallelepiped of equivalent base and 
equal altitude. 



PAEALLELEPIPEDS 333 

EXERCISE 87 

1. Find the rati<» of two rectangular parallelepipeds, If their 
dimensions are •">. I. 5, and 9, 8, 1<> respectively. 

2. Find the ratio of two rectangular parallelepipeds, if their 

altitudes are each <*» in., and their bases 5 in. by 1 in., and 1<> in. 
by 8 in. respectively. 

3. Find the volume of a rectangular parallelepiped 2 It. 

(') in. long, 1 ft. s in. wide, and 1 ft. 6 in. high. 

4. Find the volume of a rectangular parallelepiped win 

base is 27 sq. in. and whose altitude is 13| in. 

5. The volume of a rectangular parallelepiped is 1152 

cu. in. and the area of the Last- is half a square foot. Find 
the altitude. 

6. The volume of a rectangular parallelepiped with a square 

base is 'J7'-kS cu. in. and the altitude is 5 in. Find the dimen- 
sions. 

7. A rectangular tank full of water i^ 7 ft. ."> in. long h\ 
4 ft. G in. wide. How many cubic feel of water must he drawn 
<>\\' in order that the surface may he lowered a fool '.' 

8. Find to two decimal places tin- length of each side of a 
cubic reservoir that will contain exactly a gallon i-'.'A cu. in. i. 

9. .V box has as its internal dimensions IS in.. 9| in., and 
4] in. The box and cover are made of steel | in. thick, [f steel 
weighs 490 11>. per cubic font, what is the weight of the box? 

10. A steel rod 4 ft. 8 in. long is - in. wide and ]\ in. thick. 
How much does it weigh, at 490 11». per cubic l"<»t '.' 

11. If 3 eu. in. of gold beaten into ,u r <>ld leaf will o- 
75,000 aq. in. of surface, find the thickness of the leaf. 

12. The sum of the squares on tin- four diagonal a par- 
allelepiped is equivalent t<> the sum of tl on the 

twelve edges. 



334 



BOOK VII. SOLID GEOMETRY 



Proposition XL Theorem 

538. The volume of a triangular prism is equal to the 
product of its base by its altitude. 

D' 




Given the triangular prism ABC-B', with volume v> base 6, and 
altitude a. 

To prove that v = ba. 

Proof. Upon the edges AB, BC, BB' construct the parallele- 
piped ABCD-B'. 

Then ABC-B' = J ABCD-B'. § 527 

The volume of ABCD-B' = ABCD x a. § 536 

But ABCD = 2b. §126 

.'. v = ^(2 ba) = ba, by Ax. 9. Q.E.D. 

EXERCISE 88 

Find the volumes of the triangular prisms whose bases and 
altitudes are as follows : 



1. 17 sq. in., 8 in. 

2. 15.75 sq. ft., 3 ft. 

3. 3i sq. ft., 1 ft. 8 in. 

4. 51 sq. ft., 2 ft. 9 in. 

5. 15.84 sq. ft., 3 ft. 10 in. 



6. 16f sq. in., 2| in. 

7. 221 S q. in., 41 in. 

8. 33i sq. in., !\ in. 

9. 42| sq. in., 3| in. 
10. 27f sq. in., 3| in. 



11. 12 sq. ft. 75 sq. in., 2 ft. 7 in. 



PRISMS 



Pkoposition XII. Theorem 



335 



539. Tin volutin of any prism is equal to tin product 

of its /"/» by its alt if ml, . 




Given the prism AC with volume v, base b } and altitude a. 
To prove thai v = ba. 

Proof. It is possible to divide any prism in general into 
what kind of simpler prisms '.' 

1 [ow is this done '.' 

What is the volume of each of these simpler prisn 538 

What is the sum of the volumes of these simpler prisms '.' 

Whal is the sum of their bases '.' 

How does the common altitude of these simpler prisms 
compare with a, the altitude of the given prism '.' 

What conclusion can be drawn from these statements? 

Write the proof in full. 

540. Corollary 1. Prisms having equivalent bases are to 
each other as their altitudes; prisms having equal altitudes 
an to each other as their ba% 

Write thr proof in full. 

541. Corollary 2. Prisms having equivalent vnd 
,'i>i,il altitudes an equivalent, 

Writr the proof in full. 



336 BOOK VII. SOLID GEOMETRY 

EXERCISE 89 

1. If the length of a rectangular parallelepiped is 18 in., the 
width 9 in., and the height 8 in., find the total area of the surface. 

2. Find the volume of a triangular prism, if its height is 
15 in. and the sides of the base are 6 in., 5 in., and 5 in. 

3. Find the volume of a prism whose height is 15 ft., if 
each side of the triangular base is 10 in. 

4. The base of a right prism is a rhombus of which one 
side is 20 in., and the shorter diagonal 24 in. The height of 
the prism is 30 in. Find the entire surface and the volume. 

5. How many square feet of lead will be required to line an 
open cistern which is 4 ft. 6 in. long, 2 ft. 8 in. wide, and con- 
tains 42 cu. ft.? 

6. An open cistern 6 ft. long and 4^ ft. wide holds 108 
cu. ft. of water. How many square feet of lead will it take 
to line the sides and bottom ? 

7. One edge of a cube is e. Find in terms of e the surface, 
the volume, and the length of a diagonal of the cube. 

8. The diagonal of one of the faces of a cube is d. Find in 
terms of d the volume of the cube. 

9. The three dimensions of a rectangular parallelepiped are 
a, b, c. Find in terms of a, b, and c the volume and the area of 
the surface. 

10. Find the volume of a prism with bases regular hexagons, 
if the height is 10 ft. and each side of the hexagons is 10 in. 

11. An open cistern is made of iron \ in. thick. The inner 
dimensions are : length, 4 ft. 6 in. ; breadth, 3 ft. ; depth, 2 ft. 
6 in. What will the cistern weigh when empty ? when full of 
water ? (A cubic foot of water weighs 62i lb. Iron is 7.2 times 
as heavy as water ; that is, the specific gravity of iron is 7.2.) 




PYRAMIDS 337 

542. Pyramid. A polyhedron of which one face, called the 
base, is a polygon of any Dumber of Bides and the other faces 
are triangles having a common 

vertex is called a pyramid. 

The triangular fares having a 
common vertex are called the lateral 
/wis. their intersections are called 
the int> /•"/ edges, ami their common 
vertex is called tli' ex of the 
pyramid. The base of a pyramid 
may he any kind of a polygon, but 
usually a convex polygon is taken. 

543. Lateral Area. The sum <>1 the mens of the lateral taces 
of a pyramid is called the lateral area of the pyramid. 

544. Altitude. The perpendicular distance from the vertex 
t»t the plane of the base is called the altitude of the pyramid. 

545. Pyramids classified as to Bases. Pyramids are said to 

l>c triangular, quadrangular, and so on, according as their 

bases are triangles, quadrilaterals, ami s<» on. 

A triangular pyramid has four triangular faces ami is called a tetra- 
hedron. Any one of its faces may he taken as the base. 

546. Regular Pyramid. If the base 

of a pyramid is a regular polygon 

whose center coincides with the foot 

of the perpendicular let fall from the 

vertex to the base, the pyramid is 

called a regular pyramid. 

A regular pyramid is also called a right 
pyramid. 

547. Slant Height of a Regular Pyramid. The altitude of 
any one of the lateral Paces of a regular pyramid, drawn 
from the vertex <>t' the pyramid, is called the slant height. 

The slant height Lb the same whatever face La taken I ■>■ < >nly a 
regular pyramid can have a slant height. 




338 



BOOK VII. SOLID GEOMETRY 





548. Properties of Regular Pyramids. Among the properties 
of regular pyramids the following are too evident to require 
further proof than that referred to below : 

(1) The lateral edges of a regular 'pyramid are 
equal (§ 439). 

(2) The lateral faces of a regular pyramid are 
congruent isosceles triangles (§ 80). 

(3) The slant height of a regular pyramid is 
the same for all the lateral faces (§ 439). 

549. Frustum of a Pyramid. The portion of a pyramid in- 
cluded between the base and a section parallel to the base is 
called a frustum of 
a pyramid. 

The base of the pyra- 
mid and the parallel 
section are called the 
bases of the frustum. 

A more general term, 
including frustum as a special case, is truncated pyramid, the portion of 
a pyramid included between the base and any section made by a plane 
that cuts all the lateral edges. This term is little used. 

550. Altitude of a Frustum. The perpendicular distance 
between the bases is called the altitude of the frustum. 

E.g. C'C is the altitude of the frustum in the above figure. 

551. Lateral Faces of a Frustum. The portions of the lateral 

faces of a pyramid that lie between the bases of a frustum are 

called the lateral faces of the frustum. 

In the case of a frustum of a regular pyramid the lateral faces are 
congruent isosceles trapezoids. The sum of the areas of the lateral faces 
is called the lateral area of the frustum. 

552. Slant Height of a Frustum. The altitude of one of the 
trapezoid faces of a frustum of a regular pyramid is called the 
slant height of the frustum. 

Thus JOT in the above figure is the slant height. 



PYRAMIDS 



■ 



Pboposihoh Mil. Theorem 

553. Tin lateral area of a regular pyramid is equal 
to half th product of its slant height by tJu perimeter 
of its bo 




Given the regular pyramid V-ABCDE, with / the lateral area, 
s the slant height, and p the perimeter of the base. 



gruenl 



To prove that / = .\ sp. 

Proof. The .1.1/:. VBC, VCD, VDE, and YEA are ron- 

548 

The area of each A= \& x its ba 

The sum of the I • of the triangles =p 

.-. the sum of the areas of these A = }, sp 

But the sum of the areas of these A = /. 

.*. I = \ >/'. by Ax. 

554. Corollabt. The lateral area 
. rust, mi of a regular pyramid 
equal to half the sum of the perim- 
,-.< ,,,' the bases multiplied by t/u jf 
slant height of th frustum. 

How ia the an oid found (| 

ru.nT ? What ia the sum of their !"■ 1 their a] : 

What is the sum of their areas I the formu] 



325 

\ . 11 

\-.. 1 

543 

Q.E.D. 




B 



340 BOOK VII. SOLID GEOMETRY 

Proposition XIV. Theorem 

555. If a pyramid is cut by a plane parallel to the 
base : 

1. The edges and altitude are divided proportionally \ 

2. The section is a polygon similar to the base. 



Given the pyramid V-ABCDE cut by a plane parallel to its base, 
intersecting the lateral edges in A\ B\ C\ D\ E\ and the alti- 
tude VO in O'. - ' 

• „, VA' VB' VO' 

1. To prove that -j^— = -=-- = • • . = — -• 

Proof. Since the plane A'D' is II to the plane AD, Given 

.'.A'B' is WtoAB,B'C'is II to BC, • • ., and A'O' is II to ^4 0. §453 

VA 1 VB' VO' , . _, 

.-. = = ...= ? by §274. q.e.d. 

VA VB VO J 

2. To prove the section A 1 B'C'D'E' similar to the base AB CDE. 

Proof. Since AVA'B' is similar to AVAB, AVB'C similar 
to AVBC, and so on (why ?), how can the corresponding sides 
of the polygons be proved proportional ? 

Since A'B' is II to AB, B'C to BC, etc. (why ?), 

how can the corresponding angles be proved equal f 

Then why is A'B'C'D'E' similar to ABODE? 



PYRAMIDS 



341 



556. Cobollabt 1. Any tectum of a pyramid parallel to 
the base is to the base as the square of the distance from the 
vertex is to the %quan of the altitude of the j/t/ntmiif. 

VO r.i 



Fur 



VO 



Therefore 



YO'~ 



V A 

.\n 

~AIi 

a i; 



A ir 



But. from similar polygons, 

A'B'C'jyE' 
ABCDE 

Hence, by substituting, 

A'B'C'irE' 
ABCDE 



A'B 
IS 



V6 2 



a 



S 288 

§27" 

§334 
Ax. 8 




557. Cobollaby 2. Tf two pyramids have equal atiitu 
and equivalent bases, sections made by planes parallel to the 
,. and at equal distances from the vertices, >><: equival 



Wha1 is the ratio of A /; ' D W to ABCDE? 



A 






II- .w can this be shown to equal VO f ■ VO ' 
What is the ratio of X'Y '// to ZTZ? 






How can this bi sho wn to equ al W P* WP ? 
An- the ratio* VO**: ro*and WP*\ WF* equal? 
- ooeit Isghren that ABCDE ZrZ,whatcan be said ol A'&C 1>'E' 
and X'Y'Z'l 



342 



BOOK VII. SOLID GEOMETRY 



Proposition XV. Theorem 

558. Two triangular pyramids having equivalent 
bases and equal altitudes are equivalent. 




B B' 

Given two triangular pyramids, V-ABC and 7'-^ f B r C', having 
equivalent bases and equal altitudes. 

To prove that V-ABC and V'-A'B'C are equivalent. 

Proof. Suppose the pyramids are not equivalent, and 

V'-A'B'C' > V-ABC. 

Place the bases in the same plane, and suppose the altitude 
divided into n equal parts, calling each of these parts h. 

Through the points of division pass planes parallel to the 
base, cutting the pyramids in DEF, GHI, • • • , D'E'F', G'H'I', 

On A'B'C, D'E'F', G'H'I', and other parallel sections, if any, 
construct prisms with lateral edges parallel to A'V, and with 
altitude h. In the figure these are represented by X', Y', and Z'. 

On DEF, GHI, and other parallel sections, if any, as upper 
bases, construct the prisms Y, Z, with lateral edges parallel to 
VA, and with altitude h. 

Then since DEF = D'E'F', § 557 

and h = h, Idem 

.*. prism Y= prism Y'. § 541 
Similarly prism Z = prism Z'. 



PYRAMIDS 343 

But A' + )" + /'> I"- 17;'' '. 

and V+Z< V-ABC. Ax. 11 

.-. v'-A'B'C — V-ABC < X' + V + Z' — ( Y+Z), 
or V'-A'B'C V- iBC< ' \ . 

That is, tlic difference between the pyramids mnsl be l< 
than the difference between the 9< ta of prisms. 

Now by increasing n indefinitely, and consequently de- 
creasing // indefinitely, A' can be made less than any assigned 
quantity. 

Hence whatever difference we suppose to exist between the 
pyramids, A Van be made smaller than thai supposed difference. 

But this is absurd, since we have shown thai A' is greater 
than the difference) if any exist 

Hence it Leads to a manifest absurdity to suppose that 
I '-.17;'''; . V-ABC. 

In the same \\a\ it leads to an absurdity to suppose that 
V-ABO V'-A'B'C. 

.-. V-ABC = V'-A'B'C. . q.e.d. 

EXERCISE 90 

1. Tli" slant height of a regular pyramid is 6 in., and the 
base is an equilateral triangle of altitude 2Vo in. Find the 
lateral area of the pyramid. 

2. The slant height of a regular triangular pyramid equals 

the altitude of the base. The area of the base is ^ 3 s<j. ft. 

Find the total area of the pyramid. 

3. A pyramid has for its base a right triangle with hy- 
potenuse 5 and shorte le •"•. Another "tie of equal altitude 

has tor its base an equilateral triangle with side 2\ 2 \ •">. 
Prove the pyramids equivalent. 



344 BOOK VII. SOLID GEOMETRY 

Proposition XVI. Theorem 

559. The volume of a triangular pyramid is equal to 
one third the product of its base by its altitude. 



Given the triangular pyramid E-ABC, with volume v, base &, 
and altitude a. 

To prove that v ~\ oa ' 

Proof. On the base AB C construct a prism ABC-DEF. 

Through DE and EC pass a plane ODE. 
Then the prism is composed of three triangular pyramids 
E-ABC, E-CFD, and E-A CD. 

Now the pyramids E-CFD and E-A CD have the same altitude 
and equal bases CFD and A CD. § 126 

. ' . E-CFD = E-A CD. § 558 

But pyramid E-CFD is the same as pyramid C-DEF, 
which has the same altitude as pyramid E-ABC, 

and has base DEF equal to base ABC. § 511 

.'. E-CFD = E-ABC. § 558 

. ' . E-ABC = E-CFD = E-A CD. Ax. 8 

.'. pyramid E-ABC = ^ prism ABC-DEF. 

But the volume of ABC-DEF =ba. § 539 

.-. v = i ba, by Ax. 4. Q-E.d. 

560. Corollary. The volume of a triangular pyramid is 
equal to one third the volume of a triangular prism of the 
same base and altitude. 



PYRAMIDS 



Proposition XVII. THEOREM 



345 



561. The volunu of any pyramid is equal to om third 
tl,, product of its base by its altitudi . 




Given the pyramid V-ABCDE, with volume v } base b, and alti- 
tude a. 

To prove that v = J ba. 

Proof. Through the edge VD and the diagonals of the ba 

l>.\ , />/;, pass planes. 

These planes divide the pyramid V-ABCDE into three tri- 
angular pyramids. 

What can he said as to the altitudes of the original pyramid 
and of the triangular pyramids *.' 

What ran be Baid as to the hase of the original pyramid in 
relation to the bases of the triangular pyramids '.' 

What is the volume of each triangular pyramid? 

What is the sum of the volumes of the triangular pyramids '.' 

( omplete the proof. 

562. Corollary. The volumes of two pyramids an tch 
other as the products of their bases and altitudes; pyramids 
having equivalent bases are to ><i<h <>tfi> r ,i.< their altitudi 
pyramids having >'/i"f/ altitudes are '■> each other as their 
bases; pyramids having equivalent ba fual altitudes 

art equivalent 



346 BOOK VII. SOLID GEOMETRY 

EXERCISE 91 

Find the lateral areas of regular pyramids, given the slant 
heights and the perimeters of the bases, as follows : 

1. s — 34 in., p = 57 in. 3. s = 2 ft. 7 m.,p = 4 ft. 6 in. 

2. 5 = 8f in., p = 17| in. 4. s=127ft. 5 in., _p = 63 ft. 2 in. 

Find the lateral areas of frustums of regular pyramids, 
given the slant heights of the frustums and the perimeters of 
the bases, as follows : 

5. s = 4 in., p = 8 in., p' = 6 in. 

6. s = 5£ in., ^j> = 9| in., £>' = 7| in. 

7. s = 2 ft. 3 in., p = 4 ft. 8 in., 79' = 3 ft. 9 in. 

Find the volumes of pyramids, given the altitudes and the 
areas of the bases, as follows : 

8. a = 7 in., h = 9 sq. in. 11. a = 3| in., h = 5| sq. in. 

9. a = 6 in., & = 23 sq. in. 12. a = 4| in., 6 = 19 sq. in. 
10. a = 17 in., 6=51 sq. in. 13. a = 27.5 ft., b = 325 sq. ft. 

.Fu'wrf ^/^ lateral areas of regular pyramids, given the slant 
heights, the number of sides of the bases, and the length of 
each side, as follows : 

14. s = 2.3 in., n = 4, 1= 2.1 in. 

15. s = 3.7 in., n = 6, 1= 2.9 in. 

16. s = 5.33 in., n = 8, Z= 3 in. 

i^m^ //*<? volumes of pyramids, given the altitudes and a 
description of the bases, as follows : 

17. a = l in., the base a square with side 2 in. 

18. a = 6| in., the base a square with diagonal 3 V2 in. 

19. a = 8.9 in., the base a triangle with each side 3.7 in. 



PYRAMIDS 347 

20. Find the lateral area of a regular pyramid, if the slant 
height is 16 ft. and the base is a hexagon with side 12 ft. 

21. Find the lateral area of a regular pyramid, if the slant 
height is 8 ft. and the base is a pentagon with side 5 ft. 

22. Find the total surface of a regular pyramid, if the slant 
height is 6 it. and the base is a square with side 4 ft. 

23. Find tlif total surface of a regular pyramid, if the slant 
height is 18 ft. and the base is a square with side 8 ft. 

24. Find the total surface of a regular pyramid, if the slant 
height is 16 ft. and the base is a triangle with side 8 ft. 

25. The volume of a pyramid is 26 cu. ft. 936 cu. in. and 
each side of its square base is 3 ft. 6 in. Find the height. 

26. The volume of a pyramid is 20 cu. ft. and the sides of 
its triangular base are 5 ft,, 4 ft., and 3 ft. respectively. Find 
the height. 

27. Find the volume of a regular pyramid with a square 

base whose side is 40 ft., the lateral edge being KM t't. 

28. Find the volume of a regular pyramid wla.se slant height 
is \'J ft. and whose base is an equilateral triangle inscribed in 
a circle of radius 10 ft. 

29. Having given the base edge a and the total surface t <■!' 
a regular pyramid with a square base, hud the heiglrl //. 

30. Having -hen tin- base edge " ami the total surface t of 
a regular pyramid with a square base, find the volume v. 

31. The eight edges of a regular pyramid with a square base 

are equal ami the total surface is /. Find the edge. 

32. Find the base edge a of a regular pyramid with a square 
base, having given the height /> and the total surface t. 

33. show how to find the volume of any polyhedron by 
dividing the polyhedron into pyramids. 



348 BOOK VII. SOLID GEOMETRY 

Proposition XVIII. Theorem 

563. The frustum of a triangular pyramid is equiva- 
lent to the sum of three pyramids whose common altitude 
is the altitude of the frustum and whose bases are the 
loiver base, the upper base, and the mean proportional 
between the two bases of the frustum. 





Given the frustum of a triangular pyramid, ABC-DEF, having 
ABC, or b, for its lower base ; DEF, or b\ for its upper base ; and 
the altitude a. 



To prove that ABC-DEF = \ab + \aV + \a y/bb'. 

Proof. Through A, E, and C, and also through C, D, and E, 
pass planes dividing the frustum into three pyramids. 

Then E-ABC = £ ab, 

and C-DEF=±ab'. §559 

It therefore remains only to prove that E-ACD = i a^/bb'. 
We see by the figure that we may speak of E-ABC as C-ABE, 
and of E-A CD as C-AED. 

But C-ABE : C-AED = A ABE : A AED. § 562 

Since A. 4 BE and AED have for a common altitude the 
altitude of the trapezoid ABED, 

.'. A ABE : A AED = AB : DE. § 327 

.'. C-ABE : C-AED = AB : DE, Ax. 8 

or E-ABC : E-A CD = AB : DE. Ax. 9 



PYRAMIDS 349 

In like manner E-ACD and E-CFD have a common vertex 

E and have their bases in the same plane, .1 ( 'FD, so that 

E-ACD : E-CFD = A ACD: A CFD. § 562 

Since A A CD and ( 'FD have for a common altitude the alti- 
tude of the trapezoid .1 ( 'FD, 

.-.A ACD: A CFD — AC:DF. § 327 

.*. E-ACD: E-CFD — AC : DF. Ax. 8 

1 Jut A D£F is similar to A A EC. § 555 

.*. I/; : DE = AC:DF. §'282 

.'. E-ABC : 7-;--4 CD = AC:DF. A x. 8 

.'. /•;,!/;' ' : /•;-.! < 7; = E-4< 7> : E-< 'FD. Ax. 8 

But E-CFD is the same as C-DEF, which has been shown to 
equal ^ «&'. 

.'. J o£ : E-A CD = ]■:-. VCD: \ ab'. Ax. 9 

.-. E-ACD= V.\ oh x £ itb' § 262 

.*. 7wl£C + C-D-EF -f E-A CD= : i : afl + J o£' + i a VST'. Ax. 1 
That is, ABC-DEF— \ ab + % ab 1 + J a VST', by Ax. 9. Q.e.d. 

564. Corollary 1. !Z%e volume of a frustum of a tri- 
angular pyramid may I" expressed as -J a {b -f- b' -f- V A// ). 

For we may factor by a a. 

565. Corollary 2. 27ie volume of a frustum of any 

pyramid is equal to the sum of tin 1 volumes of three pyramid* 

"•hose common attitude is the altitude of the frustum, and 

tr/mxe bases are the lower base, the upper base, and the mean 

proportional between the l><tses of the frustum. 

Extend the faces of the frustum F, forming a pyramid P. From a 
triangular pyramid I' »»f equivalent base & and equal altitude, cut ofi 
a frustum F' of the same altitude a as F. Then P= P and F= F . 
But Fand F' have equivalent bases, and F' = £a(fr + 6' fw), Hence 
F=|a(6 + fc'+VwJ 7 ). 



350 



BOOK VII. SOLID GEOMETRY 



566. Polyhedrons classified as to Faces. A polyhedron of 
four faces is called a tetrahedron ; one of six faces, a hexahe- 
dron ; one of eight faces, an octahedron ; one of twelve faces, 
a dodecahedron ; one of twenty faces, an icosahedron. 








Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron 

567. Regular Polyhedron. A polyhedron whose faces are con- 
gruent regular polygons, and whose polyhedral angles are equal, 
is called a regular polyhedron. 

It is proved on page 351 that it is possible to have only five regular 
polyhedrons. They may be constructed from paper as follows : 






Draw on stiff paper the diagrams given above. Cut through the full 
lines and paste strips of paper on the edges as shown. Fold on the dotted 
lines, and keep the edges in contact by the pasted strips of paper. 



REGULAR POLYHEDRONS 351 

Proposition XIX. Problem 

568. To cU /' rmine the numh r of regular convex pohj- 
hedrons possible. 

A convex polyhedral angle must have at least three faces, 
and the sum of its face angles must be less than 360° (§ 495). 

1. Since each angle of an equilateral triangle is 60°, convex 
polyhedral angles may be formed by combining three, four, or 
live equilateral triangles. The sum of six such angles is 360°, 
and therefore is greater than the sum of the face angles of a 
convex polyhedral angle. Hence three regular convex polyhe- 
drons are possible with equilateral triangles for faces. 

2. Since each angle of a square is 90°, a convex polyhedral 
angle may be formed by combining three squares. The sum of 
four such angles is 360°, and therefore is greater than the sum 
of the face angles of a convex polyhedral angle. Hence one 
regular convex polyhedron is possible with squares. 

."». Sine.' each angle <»!' a regular pentagon is 108° (§ 14.Vi. a 
convex polyhedral angle may be formed by combining three 
regular pentagons. The sum of four such angles is 432°, and 
therefore is greater than the sum of the face angles of a convex 
polyhedral angle. Hence cue regular convex polyhedron is 
possible with regular pentagons. 

4. The Mini of three angles of a regular hexagon is 360°, of 
;i regular heptagon is greater than 3<>0°, and so on. 

Hence only live regular convex polyhedrons are possible. 

The regular polyhedrons are the regular tetrahedron, the 
regular hexahedron, or cube, the regular octahedron, the regular 
dodecahedron, and the regular icosahedron. q.e.f. 

It adds greatly to ;i clear understanding of the five regular poly- 
hedrons if they are constructed from paper as suggested in § ">'">7. 

Since these ><>li<ls were extensively Btudied by the pupils of Plato, the 
greal Greek philosopher, they are often called the Platonic Bodi 



352 BOOK VII. SOLID GEOMETRY 

EXERCISE 92 

Find the volumes of frustums of pyramids, the altitudes and 
the bases of the frustums being given, as follows : 

1. a = 3 in., b = S sq. in., V = 2 sq. in. 

2. a = 4J- in., b = 8^ sq. in., V = 3 sq. in. 

3. a = 3.2 in., 5 = 2 sq. in., &' = 0.18 sq. in. 

4. a = 2 ft. 6 in., & = 10 sq. ft., V = 2 sq. ft. 72 sq. in. 

5. a = 3 ft. 7 in., /> = 24 sq. ft. 72 sq. in., b' = 2 sq. ft. 

6. A pyramid 2 in. nigh, with a base whose area is 8 sq. in., 
is cut by a plane parallel to the base 1 in. from the vertex. 
Find the volume of the frustum. 

7. A pyramid 3 in. high, with a base whose area is 81 sq. in., 
is cut by a plane parallel to the base 2 in. from the base. Find 
the volume of the frustum. 

8. The lower base of a frustum of a pyramid is a square 
1 in. on a side. The side of the upper base is half that of the 
lower base, and the altitude of the frustum is the same as the 
side of the upper base. Find the volume of the frustum. 

9. The lower base of a frustum of a pyramid is a square 
3 in. on a side. The area of the upper base is half that of the 
lower base, and the altitude of the frustum is 2 in. Find to 
two decimal places the volume of the frustum. 

10. A pyramid has six edges, each 1 in. long. Find to two 
decimal places the volume of the pyramid. 

11. A regular tetrahedron has a volume 2 V2 cu. in. Find to 
two decimal places the length of an edge. 

12. The base of a regular pyramid is a square I ft. on a side. 
The slant height is s ft. Find the area of the entire surface. 

13. Consider the formula v = £ a (b + V + V^), of § 564, 
when V ±= 0. Discuss the meaning of the result. Also discuss 
the case in which b = b'. 



CYLINDERS 



353 



569. Cylindric Surface. A surface generated by a straight 
line which is constantly parallel to a fixed Btraight line, and 
touches a fixed curve not in 

the plane of the straight line, 

is called a cylindric surface^ or 

a cylindrical surface. 

The moving line is called the 
generatrix and the fixed curve the 
directrix. In the figure ABC is 
the directrix. 

570. Element. The generatrix 
in any position is called an ele- 
ment of the cylindric surface. 

571. Cylinder. A solid bounded by a cylindric surface and 
two parallel plane surfaces is called a cylinder. 





It follows, therefore, that all the elements of a cylinder are equal. 
The terms bases, lateral surface, and altitude are used as with prisms. 

572. Right and Oblique Cylinders. A cylinder whose element s 
are perpendicular to its bases is called a right cylinder; other- 
wise a cylinder is called an oblique cylinder. 

573. Section of a Cylinder. A figure formed by the intersec- 
tion of a plane and a cylinder is called ^section of the cylinder. 



354 



BOOK VII. SOLID GEOMETKY 



Proposition XX. Theorem 

574. Every section of a cylinder made by a plane 
passing through an element is a parallelogram. 





Given a cylinder AC, and a section ABCD made by a plane pass- 
ing through the element AB. 

To prove that ABCD is a parallelogram. 

Proof. Through D draw a line in the plane ABCD II to AB. 
This line is an element of the cylindric surface. § 570 

Since this line is in both the plane and the cylindric surface, 
it must be their intersection and must coincide with DC. 
Hence DC coincides with a straight line parallel to AB. 

Therefore DC is a straight line II to AB. 

Also AD is a straight line II to BC. § 453 

.*. ABCD is a parallelogram, by § 118. q.e.d. 

575. Corollary. Every section of a right cylinder made 
by a plane passing through an element is a rectangle. 

576. Circular Cylinder. A cylinder whose bases are circles is 

called a circular cylinder. 

A right circular cylinder, being generated by the revolution of a rec- 
tangle about one side as an axis, is also called a cylinder of revolution. 



CYLINDERS 

Proposition- XXI. Tiikokkm 
577. Tin bases of a cylinder are congruent, 

G C 



355 




Given the cylinder AC, with bases ABE and DCG. 
To j>r<"'r that ABE is congruent f>> DCG. 

Proof. Let A, B, E be any three points in the perimeter of 

the lower base, and AD, EC, Ed be elements of the surface. 

Draw AB, AE, El',, DC, 1><;, GC. 

Then AD, BC, EC are equal, § 571 

and parallel. § 569 

.-. AB = Dc, AE= DG, EB= GC. § 130 

.*. A A HE is congruent to A DCG. § 80 

Place the lower base on the upper base so that HlqAABE 
shall fall on the A DCG. Then A, B, E will fall on D, C, G. 

Therefore all points in either perimeter will coincide with 
points in the other, and the bases are congruent, by § 66. q. e. d. 

578. Cmkollary 1. A//// two parallel sr<-ti<>iis of a cylinder, 
cutting all the elements, are congruent 

579. Corollary 2. Any section of a cylinder parallel to 
the l>ase is congruent t<> the base. 

580. Corollary 3. The straight line joining the center* of 
tin hoses of a circular cylindt r jesses through thr centers of 
all sections of the cylindi r parallel f>> the bat 



356 



BOOK VII. SOLID GEOMETRY 



581. Tangent Plane. A plane which contains an element of 
a cylinder, but does not cut the surface, is called a tangent plane 
to the cylinder. 

582. Construction of Tangent Planes. From a consideration 
of the nature of a tangent plane and of the construction of a 
cylindric surface it is evident that : 

A plane passing through a tangent to the base of a circular 
cylinder and the element drawn through the point of contact is 
tangent to the cylinder. 

If a plane is tangent to a circular cylinder, its intersection 
with the plane of the base is tangent to the base. 

583. Inscribed Prism. A prism whose lateral edges are ele- 
ments of a cylinder and whose bases are inscribed in the bases 
of the cylinder is called an inscribed prism. 

In this case the cylinder is said to be circumscribed about the prism. 





Inscribed Prism 



Circumscribed Prism 



584. Circumscribed Prism. A prism whose lateral faces are 
tangent to the lateral surface of a cylinder and whose bases 
are circumscribed about the bases of the cylinder is called a 
circumscribed prism. 

In this case the cylinder is said to be inscribed in the prism. 



CYLINDERS 



357 



585. Right Section. \ Bection of a cylinder made by a plane 
that cuts all the elements and is perpendicular to them is called 
a right section of the cylinder. 

586. Cylinder as a Limit. From the work already done in 
connection with limits, and from the nature of the inscribed 
and circumscribed prisms, the following properties of the 
cylinder may new be assumed without further proof than 
that given below : 

If a prism whose base is a regular polygon is inscribed in or 
circumscribed about a circular cylinder^ and if the number of 
sides of the prism is indefinitely increased^ 

1. The volume of the cylinder is the limit of the volume of 
the prism. 

2. The lateral area of the cylinder is the limit of the lateral 
area of the prism. 

3. The perimeter of a right section of the cylinder is the limit 
of the perimeter of a right section of the prism. 




For as we increase the number of sides of the base of the inscribed 
or circumscribed prism whose base is a regular polygon, the perimeter 
of the base approaches the circle as its limit ($ 381). 

This brings the lateral surface of each prism nearer and nearer the 

lateral surface of the cylinder. It also brings the volume of each prism 

o-er and nearer the volume of the cylinder. In the same way it bril 
the right Bection of each prism nearer and nearer the right section of 
the cylinder. 



358 BOOK VII. SOLID GEOMETRY 

Proposition XXII. Theorem 

587. The lateral area of a circular cylinder is equal 
to the product of an element by the perimeter of a 
right section of the cylinder. 



Given a circular cylinder C, / being the lateral area, p the perim- 
eter of a right section, and e an element. 

To prove that I = ep. 

Proof. Suppose a prism with base a regular polygon to be 
inscribed in C, V being its lateral area and p' being the perim- 
eter of its right section. 

Then V = ep'. § 512 

If the number of lateral faces of the prism is indefinitely 

increased, , .. 

Z' approaches £ as a limit, 

anc [ y approaches j? as a limit, § 586 

and consequently ep' approaches ep as a limit. 

.-. l = ep, by §207. Q-e.d. 

588. Corollary. The lateral area of a cylinder of revolu- 
tion is equal to the product of the altitude by the circum- 
ference of the base. 

In the case of a right circular cylinder of altitude a, lateral area I, 
total area t, and radius of base r, we have 

I = 2 7rra, and t = 2 irra + 2 7rr 2 = 2 7rr (a + r). 



CYLINDERS 



359 



Proposition Will. Tiikokkm 
589. The volume of a circular cylinder is equal to the 

product Of its l>'/sr bif its (lltitudi . 




Given a circular cylinder C, b being the base, v the volume, 
and a the altitude. 

To prove that v = ba. 

Proof. Suppose a prism with base a regular polygon to be 
inscribed in C, />' being its base and v* being its volume. 

Then c' = h'u. §539 

If the number of lateral faces of the prism is indefinitely 

increased, 

v' approaches v as a limit, § 586 

// approaches b as a limit, § 381 
and consequently b'a approaches ba as a limit. 

But u'= b'a, whatever the number of sides. § 539 

.-. c = ba, by § 207. q.e.d. 

590. Corollary. The volume of a cylinder of revolution 
with radius r and altitude a is irr^a. 

What is the area of the base ? By what is this to be multiplied ? 

591. Similar Cylinders. Cylinders generated by the revolu- 
tion of similar rectangles about corresponding sides are called 
similar cylinders of revolution. 

§§ 601 and 592 may be omitted without destroying the sequence. 



360 



BOOK VII. SOLID GEOMETRY 



Proposition XXIV. Theorem 

592. The lateral areas, or the total areas, of similar 
cylinders of revolution are to each other as the squares 
of their altitudes or as the squares of their radii ; and 
their volumes are to each other as the cubes of their 
altitudes or as the cubes of their radii. 




Given two similar cylinders of revolution, / and V denoting their 
lateral areas, f and V their total areas, v and v f their volumes, 
a and a 1 their altitudes, and r and r' their radii. 

To prove that l:l'=t:t' = a 2 : a' 2 = r 2 : r'% 

and that v : v' = a 3 : a rd = r 3 : r'\ 

Proof. Since the generating rectangles are similar, § 591 

a r a-\-r 
'''~a 1 = ? = 'a~ r +r r § 269 

Also we have by this proportion and § 588, 



I 

V 



2 irra 
2 irr'a' 



ra 
r'a' 



a 



r 



12 



a 



i-y 



But t = 2 irra -f 2 irr 2 (§ 588), and v = irr^a 
t 2 irija + r) r(a + r) r 

'''t l ~~~~ 2 irr' (a' + r*) ' ' r'(a' + r') ' ' ? 



590 



a' 



,'2 



a 



12' 



and 



irv^a 



v inra v a r a 
v' = ^rTV = 7 2 X a' ~ 7 s == a 1 *' 



Q.E.D. 



CYLINDERS 361 

EXERCISE 93 

1. The diameter of a well is 6 it. and the water is 7 ft. 
deep. How many gallons of water are there in the well, reck- 
oning 7^- gal. to the cubic fool '.' 

2. When a body is placed under water in a right circular 
cylinder 60 centimeters in diameter, tin- level of the water ri 
40 centimeters. Find the volume of the body. 

3. I I<:>\v many cubic yards of earth must be removed in 
constructing a tunnel 100 yd. long, the section being a semi- 
circle with a radius of 18 ft. ? 

4. How many square feet of sheet iron are required to 
make a pipe 18 in. in diameter and 40 ft. long ? 

5. Find the radius of a eylindric pail 14 in. high that will 
hold exactly 2 CU. ft. 

6. The height of a eylindric vessel that will hold 20 liters 
is equal to the diameter. Find the altitude and the radius. 

7. If the total surface of a right circular cylinder is t and 
the radius of the base is r, find the altitude '/. 

8. If the lateral surface of a right circular cylinder is / 
and the volume is r, find the radius /• and the altitude a. 

9. If the circumference of the base of a right circular cyl- 
inder is c and the altitude is a, find the volume v. 

10. If the circumference of the base of a right circular 
cylinder is c and the total surface is t, find the volume v. 

11. If the volume of a right circular cylinder is y and the 
altitud ■. find the total surface t 

12. If v is the volume of a right circular cylinder in which 
altitude equals the diameter, find the altitude <i and the 

total Burfac» 

13. From the formula t = 2irr(a + >•) 588) find the value 
of r. (Omit unless quadratics have been studied.) 



362 



BOOK VII. SOLID GEOMETRY 




593. Conic Surface. A surface generated by a straight line 
which constantly touches a fixed plane curve and passes 
through a fixed point not in the plane 
of the curve is called a conic surface or 
a conical surface. 

The moving line is called the generatrix, the 
fixed curve the directrix, and the fixed point the 
vertex. 

Hold a pencil by the point and let the other 
end swing around a circle, and the pencil will 
generate a conic surface. 

We may also swing a blackboard pointer 
about any point near the middle, so that either 
end shall touch any fixed plane curve, and thus 
generate a conic surface. Such a surface is rep- 
resented in the annexed figure. 

594. Element. The generatrix in any position is called an 

element of the conic surface. 

If the generatrix is of indefinite length, the surface consists of two 
portions, one above and the other below the vertex, which are called 
the upper nappe and lower nappe respectively. The two nappes are shown 
in the above figure. 

595. Cone. A solid bounded by a conic surface and a plane 
cutting all the elements is called a cone. 

The conic surface is called 
the lateral surface of the cone, 
and the plane surface is called 
the base of the cone. 

The vertex of the conic sur- 
face is called the vertex of the 
cone, and the elements of the 
conic surface are called the ele- 
ments of the cone. 

The perpendicular distance 
from the vertex to the plane of 
the base is called the altitude of 
the cone. 




(ONKS 



:]i;:>, 



596. Circular Cone. 
circular cone. 



A cone whose base is a circle is called a 



The straight line joining the vertex of a circular cone and the center 
of the base is called the axis of the cone. 

597. Right and Oblique Cones. A circular cone whose axis is 
perpendicular to the base is called a right cone ; otherwise a 
circular cone is called an oblique cone. 

598. Cone of Revolution. Since a right 

circular cone may be generated by the 

revolution of a right triangle about one 

of the sides of the right angle, it is called 

a cone of revolution. 

In this case the hypotenuse corresponds to 
an clement of the surface and is called the slant Jieight. 

599. Conic Section. A section formed by the intersection of a 
plane and the conic surface of a cone of revolution is called a 
coiii' 1 section. 






Fig. 1 



Fig. 2 



Fig. 3 



Fig. 4 



Fig. 5 



In Fig. I the conic section is two intersecting straight lines, and this 

Liscussed in § 000. This is true for all kinds of com-. 

In Fi_ r . 2 the conic section is a circle, and this is discussed in § 601. 

In Fig. 3 the conic section is called an ellipse, the form a circle seems 
to take when Looked at obliquely. The orbit of a planet is an elli] - 

In Fig. 4 the conic section is & parabola, the path of a projectile (in a 
vacuum). Here the cutting plane is parallel to an element. 

In Fig. 5 the conic section is an hyperbola. 

The general study of conic sections is not a part of elementary geome- 
try, but the names of the sections may profitably be known. 



364 



BOOK VII. SOLID GEOMETRY 



Proposition XXV. Theorem 

600. Every section of a cone made oy a plane pass- 
ing through its vertex is a triangle. 





Given a cone, with AVB a section made by a plane passing 
through the vertex V. 

To prove that A VB is a triangle. 

Proof. AB is a straight line. § 429 

Draw the straight lines VA and VB. 

The lines VA and VB are both elements of the surface of 
the given cone. § 594 

These lines lie in the cutting plane, since their extremities 
are in the plane. • § 422 

Hence VA and VB are the intersections of the conic surface 
with the cutting plane. 

But VA and VB are straight lines. Const. 

Therefore the intersections of the conic surface and the 
plane are straight lines. 

Therefore the section AVB is a triangle, by § 28. q.e.d. 



CONES 



365 



Proposition XXVI. Theorem 

601. In a circular cone a section made by a plane 
parallel to the base is a circle. 





Given the circular cone V-ABCD, with the section A'B'C'D' 
parallel to the base. 

To prove that A'B'C'D 1 is a circle. 

Proof. Let be the center of the base, and let O 1 be the point 
in which the axis VO pierces the plane of the conic section. 

Through TO and any elements r.l. 17;. pass planes cutting 
the base in the radii OA, OB, and cutting the section A'B'C'D' 
in the straight lines O'A', O'B'. 

Then O'A' and O'B' are II respectively to OA and OB. r>:> 

Therefore the A AOV and OBV are similar respectively to 
the AA'u'V and O'B'V. § 285 

OA VO OB 



§ 282 
§ 162 



' 'O'A 1 VO' <>'B' 
But OA — OB. 

.-. oM'=07J'(§263),;nid A'B'C'D' is a rircle ? by §159. Q.e.d. 

602. < Sobollaby. The axis of a circular cone pass( s through 
t/ (t center oj 7/ secii<m which is parallel to the f"tse. 



366 BOOK VII. SOLID GEOMETRY 

603. Tangent Plane. A plane which contains an element of 
a cone, but does not cut the surface, is called a tangent plane 
to the cone. 

604. Construction of Tangent Planes. 
It is evident that : 

A plane passing through a tangent to 
the base of a circular cone and the ele- 
ment drawn through the p>oint of contact 
is tangent to the cone. |Jf 

If a pAane is tangent to a circular |jj 
cone its intersection with the plane of 
the base is tangent to the base. 

605. Inscribed Pyramid. A pyramid whose lateral edges are 
elements of a cone and whose base is inscribed in the base of 
the cone is called an inscribed pyra mid '. 

In this case the cone is said to be circumscribed about the pyramid. 






Inscribed Pyramid Circumscribed Pyramid 

606. Circumscribed Pyramid. A pyramid whose lateral faces 
are tangent to the lateral surface of a cone and whose base 
is circumscribed about the base of the cone is called a cir- 
cumscribed pyramid. 

In this case the cone is said to be inscribed in the pyramid. 



COXES 



367 




607. Frustum of a Cone. The portion of a cone included be- 
tween the base and a section parallel to the base is called a 
frustum of a cone. 

The base of the cone and the parallel section 
are together called the bases of the frustum. 

The terms altitude and lateral area of a frus- 
tum of a cone, and slant height of a frustum 
of a right circular cone, are used in substan- 
tially the same manner as with the frustum of 
a pyramid (§^ •">•">(), 551, 552). 

608. Cones and Frustums as Limits. The following proper- 
ties, similar to those of § 586, are assumed without proof : 

If a pyramid whose base is a regular polygon is inscribed in 
or circumscribed about a circular cone, and if the number of 
sides of the base of the pyramid is indefinitely increased, the 
volume <>/ the cone is the limit of the volume of the pyramid^ 
and Hie lateral area of the cone is the limit of the lateral area 
of tin' pyramid. 




The volume of a frustum of a ranr is the limit of the volumes 
of the frustums of the inscribed and circumscribed 'pyramids^ 
if the number <>f lateral faces is indefinitely increased^ on, I 
the lateral area of the frustum of a <■>>/"■ is the limit of the 
Intern/ areas of the frustums of the inscribed and circumscribed 
pyramids, the bases being regular polygons* 



368 BOOK VII. SOLID GEOMETRY 

Proposition XXVII. Theorem 

609. The lateral area of a cone of revolution is equal 
to half the product of the slant height by the circumfer- 
ence of the base. 



Given a cone of lateral area /, circumference of base c, and slant 
height s. 

To prove that 1= ~^sc. 

Proof. Suppose a regular pyramid to be circumscribed about 
the cone, the perimeter of its base beings and its lateral area V. 

Then V = \ sp. § 553 

If the number of the lateral faces of the circumscribed pyra- 
mid is indefinitely increased, 

V approaches / as a limit, § 608 

p approaches casa limit, § 381 

and consequently \ sp approaches ^ «c as a limit. 

But V = % sj), whatever the number of sides. § 553 

.-. I =i 2 sc, by § 207. q.e.d. 

610. Corollary. If I denotes the lateral area, t the total 
area, s the slant height, and r the radius of the base of a cone 
of revolution, then 

I = |(2 irr X s) = irrs ; 

t = irrs + tt^ 2 = 7rr (s + r). 



cnXES 369 

EXERCISE 94 

Find the lateral areas of cones of revolution, given the slant 
heights and the circumferences of th bases respectively as 
follows : 

* 

1. 2|in.,5fin. 4. 3.7 in., 5.8 in. 7. 2 ft. 6 in., 4 ft. 8 in. 

2. I; in., sj in. 5. 5.3 in., 9.7 in. 8. 3 ft. 7 in., 8 ft. 6 in. 

3. ('»,";. in., lo.i, J,,. 6. 6.5 in., 11.6 in. 9. 5 ft. 8 in., 12 ft. 4 in. 

Find the lateral areas of cones of revolution, given the slant 
Jieights and the radii of the bases respectively as follows: 

10. 3fin.,2£in. 13. 6.4 in., 4.8 in. 16. 2 ft. 3 in., 8 in. 

11. 2£in.,lfin. 14. 7.2 in., 5.3 in. 17. '4 ft. 6 in., 2 ft. 

12. \\ in...".] in. 15. 8.9 in^ 5.6 in. 18. 6 ft. 9 in., 3 ft. 2 in. 

Find the total areas of cones of revolution, given the slant 
heights and the radii of the bases respectively as follows: 

19. 3 in., 2 in. 21. 7 in., 4 in. 23. 6 It.. 1 it. 

20. 5 in.. 3 in. 22. 9 in., 5 in. 24. 1 2 ft., 5 ft. 

25. Deduce a formula for finding the lateral area of a cone of 
revolution in terms of the radius of the base and the altitude. 

26. Deduce a formula for finding the slant height in terms 
of the lateral area and the circumference of the base. 

27. Deduce a formula for finding the slant height in terms 
of the lateral area and the radius of the base. 

28. Deduce a formula for finding the radius of the base in 
terms of the Intend area and the slant height. 

29. Deduce a formula for finding the slant height in terms 
of the total area and the radius of the bas 

30. Deduce a formula for finding the circumference of the 

base in terms of the lateral area and the slant height. 



370 BOOK VII. SOLID GEOMETRY 

Proposition XXVIII. Theorem 

611. The volume of a circular cone is equal to one 
third the product of its base by its altitude. 



Given a circular cone of volume v, base b, and altitude a. 

To prove that v = ^ba. 

Proof. Suppose a pyramid with base a regular polygon to be 
inscribed in the cone, V being its base and v' its volume. 

Then v' = J b'a. § 561 

If the number of lateral faces of the pyramid is indefinitely 

increased, 

v' approaches v as a limit, § 608 

b' approaches b as a limit, § 381 

and consequently b'a approaches ba as a limit. 

.-. v = ±ba, by § 207. Q.e.d. 

612. Corollary. In a circular cone of radius r and alti- 
tude a, v = 1 7rr 2 a. 

Eor the area of the base is 7rr 2 (§ 389). 

613. Similar Cones. Cones generated by the revolution of 
similar right triangles about corresponding sides are called 
similar cones of revolution. 

In case § 614 is omitted this definition may also be omitted. 



CONES 371 

EXERCISE 95 

Find the volumes of circular mnrx, t/iren the <tltitvdes and 
the areas of the bases respectively as follows: 

1. 4 in., 8 sq. in. 4. 6.3 in., 3.8 sq. in. 

2. :>} in.. 9f sq. in. 5. 7.8 in., 6.9 sq. in. 

3. 5| in., 10V sq. in. 6. 9.3 in., 16.8 sq. in. 

Find the e<>/ii/nes of circular cones, f/ireii the, altitudes and 
the radii of the bases respectively as follows: 

7. 4 in., 3 in. 10. 9.8 in., 4.3 in. 

8. 6 in., 4 in. 11. 10.5 in., 0.2 in. 

9. 8 in., 5 in. 12. 14.*) in., 9.6 in. 

13. How many cnbic feet in a conical tenl 10 It. in diameter 
and 7 ft. high? 

14. How many cubic feet in a conical pile of earth 15 ft. in 
diameter and 8 ft. high ? 

15. Deduce a formula for finding the altitude of a circular 
cone in terms of the volume and the area of the base. 

16. Deduce a formula for finding the area of the base of a 
circular cone in terms of the volume and the altitude. 

17. Deduce a formula for finding the altitude of a circular 
cone in terms of the volume and the radius of the base. 

18. Deduce a formula for finding the radius of the base of 
a circular cone in terms of the volume and the altitude. 

19. Deduce a formula for finding the volume of a cone of 
revolution in terms of the slant height and the radius of the 
base. 

20. Deduce formulas for finding the slant heighl and the 

altitude of a cone of revolution in terms of the volume and the 
radius of the base. 



372 BOOK VII. SOLID GEOMETRY 

Proposition XXIX. Theorem 

614. The lateral areas, or the total areas, of two sim- 
ilar cones of revolution are to each other as the squares 
of their altitudes, as the squares of their radii, or as the 
squares of their slant heights ; and their volumes are 
to each other as the cubes of their altitudes, as the cubes 
of their radii, or as the cubes of their slant heights, 



Given two similar cones of revolution, with lateral areas / and 
/', total areas t and V, volumes v and v' y altitudes a and a', radii 
r and r\ and slant heights s and s' respectively. 

To prove that I: V = t: t r = a 2 : a' 2 = r 2 : r 1 ' 2 = s 2 : s' 2 , 
and that v : v' = a z : a' z = r s : r' s = s 8 : s'\ 

Proof. - - = - = 4^ * § § 282 > 269 

a r s s + r 



I irrs r s r 2 s 2 a 2 
Z 7 = = ir7V == ? X ^'"V 2 ^^^^ 2 " 



§610 



Jt ~ ^'(3' + r') ~ ? * s' + r' " r' 2 ~7 2 ~~ a' 2 ' * 

v I irr 2 (i r 2 a r 3 a s s s . _ _ . _. 

v' ±7rr' 2 a' r' 2 a' r' 3 a ,z s' 3 

§§ 613 and 614, like §§ 591 and 592, are occasionally demanded in 
college entrance examinations. They are not needed for any exercises 
and they may be omitted without destroying the sequence. 



« ONES 373 

Proposition XXX. Theorem 

615. Tin lateral area of a frustum of a cone of revo 
lution is equal to half the sum of the circumferences <>/' 
its bases multiplied by the slant height. 




Given a frustum of a cone of revolution, with lateral area /, 
circumferences of bases c and c\ and slant height s. 

T>> 'prove that I = \ (c + c')s. 

Proof. Suppose a frustum of it regular pyramid circum- 
scribed about the frustum of the cone, as a pyramid is cir- 
cumscribed about a cone. 

Lei the lateral area of the circumscribed frustum be /'_. and 
let p and p' be the perimeters of the basi corresponding to 
c and c' respectively. The slant height is s } the same as that 
of the frustum of the cone. 

Then /' =i, <>+/). v. §554 

If the number of lateral faces of the circumscribed frustum 
is indefinitely increased, what limits do V and p -\-p' approach '.' 
Therefore what limit does l( j> + />'<■■< approach ? 
What conclusion may be drawn, as in § 587 '.' 
( !omplete the proof. 

616. Corollary. The lateral area of a frustum of a cone 
of revolution is equal to the circumference of a section equi- 
distant from it* bases multiplied by its slant height. 

How can it 1"' proved thai J (c + <■') equals the circumference of this 
section ? How are the radii related '.' 



374 BOOK VII. SOLID GEOMETKY 

Proposition XXXI. Theorem 

617. A frustum of a circular cone is equivalent to 
the sum of three cones ivhose common altitude is the 
altitude of the frustum and ivhose bases are the loiver 
base, the upper base, and the mean proportional between 
the bases of the frustum. 




Given a frustum of a circular cone, with volume v, bases &and b\ 
and altitude a. 

To prove that v = 1 a (b + V -f y/W). 

Proof. Suppose a frustum of a pyramid with base a regular 
polygon to be inscribed in the frustum of the cone, as a pyramid 
is inscribed in a cone. 

Let v' be the volume, and let x and x' be the bases corre- 
sponding to b and b' respectively. The altitude is a, the same 
as that of the frustum of the cone. 



Then v' = £ a(x + x' + Vxx'). § 565 

If the number of lateral faces of the inscribed frustum is in- 
definitely increased, what limits do v', x, x', and xx' approach ? 
Therefore what limit does \ a(x -\- x' -\- -Vxx') approach ? 
What conclusion may be drawn ? 
Complete the proof. 

618. Corollary. In a frustum of a cone of revolution, 
r and r' being the radii of the bases, v = ^ 7ra(r 2 + r' 2 + rr'). 

For b = 7rr 2 , b' = 7rr' 2 . .-. Vw — VVr 2 x 7rr' 2 = irrr'. 



CONES 375 

EXERCISE 96 

Find the lateral areas of frustums of cones, given the cir- 
cumferences of the f>asr* •nnl th,' slant heights respectively as 
follows : 

1. r = 4 in., c' = 3 in., s = 0.5 in. 

2. c = (') in., <■' = 5 in., s = 1.4 in. 

3. r = T\m., c' = 5| in., s = 2| in. 

4. c = 23 in., c' = 18 in., s = 1 (J in. 

.Fmd to /"■" decimal places the volumes of frustums of <• <■■ ,-. 
#///■. // £fo altitudes ami the areas of the bases respectively as 

follow* : 

5. a = 3 in., b = 4^ sq. in., b' = 2 sq. in. 

6. n — 4 in., /> = S 1 , sq. in., b' = ."> sq. in. 

7. t/ = 5.], in., 5 = 16 sq. in., &' = 9 sq. in. 

8. a = 6 in., b = 17 sq. in., // = 1 1 sq. in. 

iwW to few decimal places the volumes of frustums of cones 
of revolution, given the altitudes and the radii of the baSi 8 
/•< spectively as follows : 

9. a = 4 in., r = 3 in., r' = 2 in. 

10. a = 5 in., r = 3^ in., /•' = 2J in. 

11. a = G in., r = 3.7 in., r' = 3.1 in. 

12. a = 7.V in., r = 4£ in., r' = 3£ in. 

13. Deduce a formula for finding the altitude of a frustum 
of a circular cone in terms of the volume and the areas of the 
bases. 

14. Deduce a formula for finding the altitude of a frustum of 
a cone of revolution in terms of tin- volume and the radii of the 
bases. 




376 BOOK VII. SOLID GEOMETRY 

EXERCISE 97 
Industrial Problems 

1. There is a rule for calculating the strongest beam that 
can be cut from a cylindric log, as follows : 

Erect perpendiculars MD and NB on opposite 
sides of a diameter A C, at the trisection points M a 
and N, meeting the circle in D and B. Then 
A BCD is a section of the beam. 

Calculate the dimensions, the log being 16 in. in diameter. 

2. A cylindric funnel for a steamboat is 4 ft. 3 in. in diam- 
eter. It is built up of four plates in girth, and the lap of each 
joint is 1| in. Find one dimension of each plate. 

3. A tubular boiler has 124 tubes each 3| in. in diameter 
and 18 ft. long. Required the total tube surface. Answer to 
the nearest square foot. 

4. A room in a factory is heated by steam pipes. There are 
235 ft. of 2-inch pipe and 26 ft. 3 in. of 3-inch pipe, besides 2 ft. 
8 in. of 4|-inch feed pipe. Required the total heating surface. 
Answer to the nearest square foot. 

5. A triangular plate of wrought iron | in. thick is 2 ft. 7 in. 
on each side. If the weight of a plate 1 ft. square and I in. 
thick is 5 lb., find to the nearest pound the weight of the given 
triangular plate. 

6. The water surface of an upright cylindric boiler is 2 ft. 
8 in. below the top of the boiler, and is 12.57 sq. ft. in area. 
What is the volume of the steam space ? 

7. A cylinder 16 in. in diameter is required to hold 50 gal. 
of water. What must be its height, to the nearest tenth of an 
inch, allowing 231 cu. in. to the gallon ? 

8. How many square feet of tin are required to make a 
funnel, if the diameters of the top and bottom are 30 in. and 
15 in. respectively, and the height is 25 in.? 



EXK1M ISES 



377 



holes 



9. Find to two decimal places the weight of a steel plate 
4 ft. by 3 it. 2 in. by If in., allowing 190 Lb. per cubic foot. 

10. A steel plat.- for a steamship is 5 ft. long, 3 ft. 6 in. 
wide, and \ in. thick. A porthole 10 in. in diameter is cut 
through the plate. Required the weight of the finished p] 
allowing 0.29 lb. per cubic inch. Ajaswer. to two decimal pla 

11. A cast-iron base for a column i- in the form of a frus- 
tum of a pyramid, the lower base being a square 2 ft. on a side, 
and the upper base having a fourth of the area of the lower 
base. The altitude of the frustum is «.» in. Required tin- weight 
to the nearest pound, allowing 460 lb. per cubic foot. 

12. A cylinder head for a steam 

engine has the shape shown in the 

figure, where the dimensions in 

niches are: " = 12. b = 3, c = 2, 

cZ = 6,e=3,/=£,^= |,andA= 

There are six f-inch holes for bob 

Compute the weight of the plate, 

allowing 41 lb. for the weight of a 

steel plate 1 ft. square and 1 in. thick. Answer to the nearest 

tenth of a pound. 

13. A steel beam L0 in. by 5 in., in the form here shown, is 
is ft. long. The thickness of the beam is § in. and 
th.« average thickness of the flanges is § in. Find 
the weight of the beam to the nearest pound, allow- 
ing 0.29 lb. per cubic inch. 

14. A hollow steel shaft 12 ft. long is L8 in. in 
exterior diameter and 8 in. in interior diameter, hind tin- 
weight to the nearest pound, all. .win- 0.29 lb. per cubic inch. 

15. Find the expense, a1 7<> cents a square foot, of polishing 
the curved surface of a marble column in the shape of the & 
turn of a right circular cone whose slant height is 12 ft. and the 
radii of whose bases are 3 ft. 6 in. and 2 ft. 4 in. respectively. 







-t.( 



378 BOOK VII. SOLID GEOMETRY 

EXERCISE 98 
Miscellaneous Problems 

1. The slant height of the frustum of a regular pyramid is 
25 ft., and the sides of its square bases are 54 ft. and 24 ft. 
respectively. Find the volume. 

2. If the bases of the frustum of a pyramid are regular 
hexagons whose sides are 1 ft. and 2 ft. respectively, and the 
volume of the frustum is 12 cu. ft., find the altitude. 

3. From a right circular cone whose slant height is 30 ft., 
and the circumference of whose base is 10 ft., there is cut off 
by a plane parallel to the base a cone whose slant height is 
6 ft. Find the lateral area and the volume of the frustum. 

4. Find the difference between the volume of the frustum 
of a pyramid whose altitude is 9 ft. and whose bases are 
squares, 8 ft. and 6 ft. respectively on a side, and the volume 
of a prism of the same altitude whose base is a section of the 
frustum parallel to its bases and equidistant from them. 

5. A Dutch stone windmill in the shape of the frustum of a 
right cone is 12 meters high. The outer diameters at the bottom 
and the top are 16 meters and 12 meters, the inner diameters 
12 meters and 10 meters. How many cubic meters of stone 
were required to build it ? 

6. The chimney of a factory has the shape of a frustum of a 
regular pyramid. Its height is 180 ft., and its upper and lower 
bases are squares whose sides are 10 ft. and 16 ft. respectively. 
The flue throughout is a square whose side is 7 ft. How many 
cubic feet of material does the chimney contain ? 

7. Two right triangles with bases 15 in. and 21 in., and 
with hypotenuses 25 in. and 35 in. respectively, revolve about 
their third sides. Find the ratio of the total areas of the solids 
generated and find their volumes. 



EXERCISFS 379 

EXERCISE 99 

Equivalent Solids 

1. A cube each edge of which is 12 in. is transformed into 
a right prism whose base is a rectangle K'> in. long and 12 in. 
wide. Find the height of the prism and the difference between 

its total area and the total area of the cube. 

2. The dimensions of a rectangular parallelepiped are a, b } c. 
Find the height of an equivalent right circular cylinder, 
having a for the radius of its base; the height of an equivalent 

right circular cone having a for the radius of its base. 

3. A regular pyramid 12 ft. high is transformed into a regu- 
lar prism with amequivalent base. Find the height of the prism. 

4. The diameter of a cylinder is 14 ft. and its height s ft. 
Find the height of an equivalent righl prism, the base of which 
is a square with a side 4 ft. long. 

5. If one edge of a cube is e, whal is the height A of an 
equivalent right circular cylinder whose radius is ,■'.' 

6. The heights of two equivalent right circular cylinders 
are in the ratio 4:9. If the diameter of the first is 6 ft., 
what is the diameter of the second *.' 

7. A right circular cylinder 6 ft. in diameter is equivalent 
to a right circular cone 7 ft. in diameter. If the height of the 
cone is 8 ft., what is the height of the cylinder '! 

8. The frustum of a regular pyramid ft. high lias for bases 
squares 5 ft. and 8 ft. on a side. Find the height of an equiva- 
lent regular pyramid whose base is a square 1- ft. on a side. 

9. The frustum of a cone of revolution is 5 ft. high and 
the diameters of its bases are 2 ft. and 3 ft. respectively. Find 

the height of an equivalent right circular cylinder whose base 
is equal in area to the section of the frustum made by a plane 
parallel to the bases and equidistant from them. 



380 BOOK VII. SOLID GEOMETRY 

EXERCISE 100 
Review Questions 

1. Define polyhedron. Is a cylinder a polyhedron ? 

2. Define prism, and classify prisms according to their bases. 

3. How is the lateral area of a prism computed ? Is the 
method the same for right as for oblique prisms ? 

4. Define parallelepiped ; rectangular parallelepiped ; cube. 
Is a rectangular parallelepiped always a cube ? Is a cube 
always a rectangular parallelepiped ? 

5. Distinguish between equivalent and congruent solids. 
Are two cubes with the same altitudes always equivalent ? 
always congruent? Is this true for parallelepipeds? 

6. What are the conditions of congruence of two prisms ? 
of two right prisms ? of two cubes ? 

7. The opposite angles of a parallelogram are equal. What 
is a corresponding proposition concerning parallelepipeds ? 

8. How do you find the volume of a parallelepiped ? What 
is the corresponding proposition in plane geometry ? 

9. How do you find the volume of a prism ? of a cylinder ? 
of a pyramid ? of a cone ? 

10. Define pyramid. How many bases has a pyramid ? Is 
there any kind of a pyramid in which more than one face 
may be taken as the base ? 

11. How do you find the lateral area of a pyramid ? of a right 
cone ? of a frustum of a pyramid ? of a frustum of a right cone ? 

12. How many regular convex polyhedrons are possible ? 
What are their names ? 

13. Given the radius of the base and the altitude of a cone 
of revolution, how do you find the volume ? the lateral area ? 
the total area ? 



BOOK VIII 

THE SPHERE 

619. Sphere. A solid bounded by a surface all points of 

which are equidistant from a point within is called a sphere. 

The point within, from which all points on the surface are equally 
distant, is called tin' center. The surface is called the spherical surface, 
and sometimes the sphere. Half of a sphere is called a hemisphere. The 
terms radius and diameU r are used as in the case of a circle 

620. Generation of a Spherical Surface. By the definition of 

sphere it appears that a spherical surface may be generated by 

the revolution of a semicircle about its diameter as an axis. 

Thus, if the semicircle ACB revolves about AB, a spherical surface is 
generated. It is therefore assumed that a sj>htr< may bv described with 
una 'ilc n point as 'i center and any given line as a radius. 




m 




621. Equality of Radii and Diameters. It follows that: 
All radii of the same sphere >>>■<■ equal, and "II diameters of 
t/,r same sphere are equal. 

Equal spheres have equal radii, and spheres having equal 

radii "/■<■ > qual. 



382 



BOOK VIII. SOLID GEOMETRY 



Proposition I. Theorem 

622. Every intersection of a spherical surface by a 
plane is a circle. 





Given a sphere with center 0, and ABD any section of its 
surface made by a plane. 

To prove that the section ABD is a circle. 

Proof. Draw the radii OA, OB, to any two points A, B, in 
the section, and draw OC _L to the plane of the section. 

Then in A OCA and OCB, A OCA and OCB are rt. A, § 430 

OC is common, and OA = OB. § 621 

.'.A OCA is congruent to A OCB. § 89 

.*. CA = CB. § 67 

.*. any points A and B, and hence all points, in the section are 
equidistant from C, and ABD is a O, by §159. q. e.d. 

623. Corollary 1. The line joining the center of a sphere 
and the center of a circle of the sphere is perpendicular to the 
plane of the circle. 

624. Corollary 2. Circles of a sphere made by planes 
equidistant from the center are equal; and of two circles made 
by planes not equidistant from the center the one made by the 
plane nearer the center is the greater. 



PLANE SECTIONS 383 

625. Great Circle. The intersection of a spherical surface by 
a plane passing through the miter is called a §reat circle of 
the sphere. 

626. Small Circle. The intersection of a spherical surface by 
a plane which does nol pass through the center is called a 
small circle of the sphere. 

627. Poles of a Circle. If a diameter of a sphere is perpen- 
dicular to the plane of a circle of the sphere the extremities 

are railed the jioles of the circle. 

628. Corollary 1. Parallel circles have flw same poles. 

629. Corollary 2. ^47/ great circles of a sphere are equal. 

630. (iirollary 3. Every great circle bisects the spherical 
surface. 

631. Corollary 4. Two great circles bisect each other. 

The intersection of the planes passes through what point ? 

632. Corollary 5. If the planes of two great circles are 
perpendicular, each circle passes through the poles of the other. 

Draw the figure and state the reason. 

633. Corollary 6. Through two given points on the sur- 
face of a sphere a a '//-<■ of a great circle wag always be drawn. 

Do these two points, together with the center of the sphere, generally 
determine a plane? Consider the special case in which the two points 
are ends of a diameter. 

634. Corollary 7. Through three given points on the sur- 
face of a sphere one circle and "///// one can be tlnurn. 

H<>w many points determine a plane ? 

635. Spherical Distance. The length of the smaller arc of 
the great circle joining two points on the surface of a sphere 
is called the spherical distance between the points, or, where 
no confusion is likely to arise, simply the distant 



384 



BOOK VIII. SOLID GEOMETRY 



Proposition II. Theorem 



636. The spherical distances of all points on a circle 
of a sphere from either pole of the circle are equal. . 



;;||j|||j||||i! 





Given P, P\ the poles of the Circle ABC, and A, 2?, C, any points 
on the circle. 

To prove that the great-circle arcs PA, PB, PC are equal. 

Proof. The straight lines PA, PB, PC are equal. § 439 

Therefore the arcs PA, PB, PC are equal, by § 172. q.e. d. 

In like manner, the great-circle arcs P'A, P'B, P'C may be proved 
equal. 

637. Polar Distance. The spherical distance from the nearer 
pole of a circle to any point on the circle is called the polar 
distance of the circle. 

The spherical distance of a great circle from either of its poles may 
be taken as the polar distance of the circle. 

638. Quadrant. One fourth of a great circle is called a 
quadrant. 

639. Corollary 1. The polar distance of a great circle is 
a quadrant. 

640. Corollary 2. The straight lines joining points on a 
circle to either pole of the circle are equal. 



PLANE SECTIONS AND TANGENT PLANKS 385 



Proposition III. Theorem 

641. .1 point on a sphere, which is at the distance of 
a quadrant from each of two other points, not the ex- 
tremities of a diameter, is a poh of th great circL 
passing through thesi points. 





Given a point P on a sphere, PA and PB quadrants, and ABC the 
j^reat circle passing through A and B. 

To {>/•">■<■ that P is the poL of OABC. 

Proof. What kind of angles an.* tin- .J A<>P and BOP? 

How is PO related to the plane of OABC ? 

Does this prove that P is the pole of OABC ? 

642. Describing Circles on a Sphere. This proposition proves 
that we may describe a great circle on a sphere of a given radius 
so that it shall pass through two given points. 

Open the compasses the length of chord PA = Vr- + r 2 = r\2. 

643. Tangent Lines and Planes. A line or plane that has one 
point and only one point in common with a sphere, however 
far produced, is said to be tangent to the sphere, and the sphere 
to be tangent to the line or plane. 

644. Tangent Spheres. Two spheres whose surfaces have one 
point and only one point in common are said to be tangent. 



386 



BOOK VIII. SOLID GEOMETRY 



Proposition IV. Theorem 

645. A 'plane perpendicular to a radius at its extrem- 
ity is tangent to the sphere. 





Given the plane MN perpendicular to the radius OA at A. 

To prove that MN is tangent to the sphere. 

Proof. Let P be any point except A in MN. 

Then which is longer, OP or OA, and why ? 
Therefore, is P inside, on, or outside the sphere, and why ? 
What does this tell us concerning all points, except A, 
onMN? 

How, then, do we know that MN is tangent to the sphere ? 

646. Corollary. A plane tangent to a sphere is perpen- 
dicular to the radius drawn to the point of contact. 

What are the proposition and corollary of plane geometry corre- 
sponding to §§ 645 and 646 ? Do they suggest the proof of this corollary ? 

647. Inscribed Sphere. If a sphere is tangent to all the faces 
of a polyhedron, it is said to be inscribed in the polyhedron, 
and the polyhedron to be circumscribed about the sphere. 

648. Circumscribed Sphere. If all the vertices of a polyhedron 
lie on a spherical surface, the sphere is said to be circumscribed 
about the polyhedron, and the polyhedron to be inscribed 
in the sphere. 



PLANE SECTIONS AND TANGENT PLANES 387 



Proposition V*. Theorem 

649. A sphere may be inscribed in any given tetra- 
hedron. 

A 





Given the tetrahedron ABCD. 

To prove that a sphere may be inscribed in ABCD. 

Proof. Bisect the dihedral A at the edges AB, BC, and CA 
by the planes OAB, OBC, and OCA respectively. 

Every point in the plane OAB is equidistant from the fai 
ABC andABD. § 479 

Pot a like reason every point in the plane OB< ' is equidistant 
from the faces ABC and BBC ; and every point in the plane 
OCA is equidistant from the faces ABC and ADC. 

Therefore the point 0, the common intersection of these 
three planes, is equidistant from the four faces of the tetra- 
hedron and is the center of the sphere inscribed in the tetra- 
hedron, by § 647. Q e.d. 

Discussion. What is the corresponding proposition in plane geom* - 
try ? Is the line of proof similar ? 

It is shown in plane geometry that the three Lines which bisect the 
three angles of a triangle meet in a point. What is the corresponding 
proposition with reference to planes in a tetrahedron ".* Is it substan- 
tially proved in this proposition ? 

It is proved in plane geometry that a circle may be inscribed in what 
kind of a polygon ? What corresponding proposition may be interred in 
solid geometry ? 



388 BOOK VIII. SOLID GEOMETRY 

Proposition VI. Theorem 

650. A sphere may be circumscribed about any given 
tetrahedron. 




Given the tetrahedron ABCD. 

To prove that a sphere may be circumscribed about ABCD. 

Proof. Let P, Q respectively be the centers of the circles 
circumscribed about the faces ABC, ABD. 

Let PR be _L to the face ABC, and QS _L to the face ABD. 

Then PR is the locus of a point equidistant from A, B, C, 
and QS is the locus of a point equidistant from A, B, D. § 442 

Therefore PR and QS lie in the same plane, the plane J_ to 
AB at its mid-point. § 443 

If QS were II to PR, it would be _L to the face ABC. § 445 

But this is impossible, for QS is J_ to the face ABD which 
intersects the face ABC. Given 

Since Pit and QS cannot be II, and since they lie in the same 
plane, they must therefore meet at some point 0. 

.' . is equidistant from A, B, C, and D, 

and is the center of the required sphere, by § G48. q. e. d. 

651. Corollary. Through four points not in the same 

plane one spherical surface and only one can be passed. 

The center of any sphere whose surface passes through the four 
points must be in the planes mentioned in the proof, and since there is 
only one point of intersection, there can be only one sphere. 



PLANE SECTIONS AND TANGENT PLANES 389 

Proposition VII. Theorem 

652. The intersection of two spherical surfaces is a 
circle whose plane is perpendicular to the line which 
joins the centers of th spheres and whose center is in 
that line. 



Given two intersecting spherical surfaces, with centers O and 0' . 

To prove that the spherical surfaces intersect in a circle 
whose plane is /» rpendicular to <><)', and whose center is in 00'. 

Proof. Let the two great circles formed by any plane 
through and 0' intersect in .1 and B. 

Then 00' is a _L bisector of AB. § 195 

If this plane revolves about 00', the circles generate the 
spherical surfaces, and .4 describes their line of intersection. 

But during the revolution AC remains constant in length 
and perpendicular to 00 % . 

Therefore A generates a circle with center C, whose plane is 
perpendicular to 00', by § 432. Q-e.d. 

653. Spherical Angle. The opening between two great-circle 
arcs that intersect is called a spherical angle. A spherical angle 
is considered equal to the plane angle formed by the tangente 
to the arcs at their point of intersection. 

Draw a figure Illustrating this definition. 

In elementary geometry we do not consider angles formed by arcs 
of small circles. 



390 BOOK VIII. SOLID GEOMETRY 

EXERCISE 101 

1. The four perpendiculars erected at the centers of the 
circles circumscribed about the faces of a tetrahedron meet 
in the same point. 

2. The six planes perpendicular to the edges of a tetra- 
hedron at their mid-points intersect in the same point. 

3. The six planes which bisect the six dihedral angles of a 
tetrahedron intersect in the same point. 

4. Circles on the same sphere having equal polar distances 
are equal. 

5. Equal circles on the same sphere have equal polar dis- 
tances. 

6. Eind the locus of a point in a plane at a given distance 
from a given point. Also of a point in a three-dimensional space. 

7. A line tangent to a great circle of a sphere lies in the 
plane tangent to the sphere at the point of contact. 

8. An}- line in a tangent plane drawn through the point of 
contact is tangent to the sphere at that point. 

9. One plane and only one plane can be passed through a 
given point on a given sphere tangent to the sphere. 

10. Eind a point in a plane equidistant from two intersecting 
lines in the plane, and at a given distance from a given point 
not in the plane. Discuss the solution. 

11. How many points determine a straight line? a circle? 
a spherical surface ? Prove that two spherical surfaces coin- 
cide if they have this number of points in common. 

12. If two planes which intersect in the line AB touch a 
sphere at the points C and D respectively, the line CD is 
perpendicular to AB in the sense mentioned in the discussion 
under § 450, — that a plane can be passed through CD per- 
pendicular to AB. 



PLANE SECTIONS AND TANGENT FLANKS 391 



Proposition VIII. Theobem 

654. A spherical angle is measured Inj the arc of the 
great circle described from Its vertex as a pole and 
included between its sides, produced if necessary. 





Given PA and PB, arcs of great circles intersecting at P ; PA 1 
and PB\ the tangents to these arcs at P; AB, the arc of the great 
circle described from P as a pole and included between PA and PB. 

To p rove that the spherical A A I'll is measured by arc AB. 

§185 

§213 

§05 



Proof. 



In the plane POD, PB 1 is _L to PO, 
and OB is _L to PO. 
.'. PB' is I! to OB. 
Similarly PA 1 is II to OA. 

.'.ZA'PJJ' = ZAOB. § h">l 

But /. A OB is measured by arc AB. §213 

.\Z A'PB' is measured by arc -I /•'. 
,\Z -1/v; is measured by arc .1 B, by § 653. q.e.d. 

655. Corollary 1. A spherical angle has the same meas- 
ure as the dihedral angle formed by the planes of the two circles. 

656. Corollary 2. All arcs of great circles drawn through 
the pole of a given great circle arc j>> rpendicular to the given 

circle. 



392 BOOK VIII. SOLID GEOMETRY 

657. Spherical Polygon. A portion of a spherical surface 

bounded by three or more arcs of great circles is called a 

spherical polygon. 

The bounding arcs are called the sides of the polygon, the angles 
between the sides are called the angles of the polygon, and the points 
of intersection of the sides are called the vertices of the polygon. 

658. Relation of Polygons to Polyhedral Angles. The planes 

of the sides of a spherical polygon form a polyhedral angle 

whose vertex is the center of the sphere, whose face angles are 

measured by the sides of the polygon, and whose dihedral angles 

have the same numerical measure as the angles of the polygon. 

Thus the planes of the sides of the polygon 
ABCD form the polyhedral angle O-ABCD. 
The face angles BOA, COB, and so on, are 
measured by the sides AB, BC, and so on, 
of the polygon. The dihedral angle whose 
edge is OA has the same measure as the 
spherical angle BAD, and so on. 

Hence from any property of polyhedral angles ive may infer 
an analogous property of spherical polygons ; and conversely. 

659. Convex Spherical Polygon. If a polyhedral angle at the 

center of a sphere is convex (§ 491), the corresponding spherical 

polygon is said to be convex. 

Every spherical polygon is assumed to be convex unless the contrary 
is stated. 

660. Diagonal. An arc of a great circle joining two non- 
consecutive vertices of a spherical polygon is called a diagonal. 

661. Spherical Triangle. A spherical polygon of three sides 

is called a spherical triangle. 

A spherical triangle may be right, obtuse, or acute. It may also be 
equilateral, isosceles, or scalene. 

662. Congruent Spherical Polygons. If two spherical polygons 
can be applied, one to the other, so as to coincide, they are said 
to be congruent 




SPHERICAL POLYGONS 



393 



Proposition IX. Theorem 

663. Each side of a spherical triangle is less than the 
sum of the other two sides. 




Given a spherical triangle ABC, CA being the longest side. 

To prove that CA <AB + BC. 

Proof. In the corresponding trihedral angle O-ABC, 

Z COA is less than Z BOA + ZCOB. § 494 

.-. CA <AB + BC, by § 658. Q.e.d. 



Proposition X. Theorem 

664. The sum of the sides of a spherical polygon is 
less than 360°. 




Given a spherical polygon ABCD. 

To prove that AB + BC + CD + DA < 360°. 

Proof. In the corresponding polyhedral angle O-ABC D, 

Z B( ' I + Z COB + Z DOC + Z DO A < 360°. § 495 
.\ AB + BC + CD -h DA < 360°, by §658. Q.e.d. 




394 BOOK VIII. SOLID GEOMETRY 

665. Polar Triangle. If from the vertices of a spherical tri- 
angle as poles arcs of great circles are described, another 
spherical triangle is formed which is called the polar triangle 
of the first. 

Thus, if A is the pole of the arc of the great 
circle B'C\ B of C'A\ C of A'B\ A'B'C is the 
polar triangle of ABC. 

If, with A, B, C as poles, entire great circles 
are described, these circles divide the surface of 
the sphere into eight spherical triangles. 

Of these eight triangles, that one is the polar 
of ABC whose vertex A\ corresponding to A, 
lies, on the same side of BC as the vertex A ; and similarly for the other 
vertices. 

EXERCISE 102 

1. To bisect a given great-circle arc. 
What must be done to the angle at the center ? 

2. If two great-circle arcs intersect, the vertical angles are 
equal. 

3. To describe an arc of a great circle through a given point 
and perpendicular to a given arc of a great circle. 

4. Every point lying on a great circle which bisects a given 
arc of another great circle at right angles is equidistant (§ 635) 
from the extremities of the given arc. 

5. Two sides of a spherical triangle are respectively 82° 
47' and 67° 39'. What is known concerning the number of 
degrees in the third side ? 

6. Three sides of a spherical quadrilateral are respectively 
86° 29', 73° 47', and 69° 54'. What is known concerning the 
number of degrees in the fourth side ? 

7. Draw a picture of a sphere, and of an equilateral spherical 
triangle on the sphere, each side being 90°. Then draw a pic- 
ture of the polar triangle. 



SPHERICAL POLYGONS 



395 



Pkoposition XI. Theorem 

666. If one sj,]n ri< <il triangl is the polar triangle of 
another, then reciprocally tin second is the polar tri- 
angle of the first 





Given the triangle ABC and its polar triangle A'B'C 

To prove that ABC Is the polar triangle of A'B'C. 

Proof. Since -1 is the pole of B'C, 

and • C is the pole of A 'B', §665 

.\ B' is at a quadrant's distance from A and C. § 639 
.-. B' is the pole of arc AC. § 641 

Similarly A' is the pole of BC, 

and C is the pole of AB. 

.-. ABC is the polar triangle of A'B'C, by § 00.5. q.e.d. 

Discussion. Is it necessary that one of the triangles should be wholly 
within the other :' Draw the figures approximately, without using instru- 
ments, starting with A ABC having A 11 = 100°, AC = 100°, BC = 30°. 

Also draw The figures having AB = 120°. AC = 80°, BC = 40°. 

Also draw the figures suggested in Ex. 7. on page 394, where AB = 
BC = CA = 90°. Consider the proposition with these figures. 

The proposition may also be considered by starting with A ABC as 
the polar triangle of AA'l'.'C, and proving that A A'B'C' is the polar 
triangle of A ABC. 

It is desirable in the study of spherical triangles to have a spherical 
blackboard. Where this is not available, any wooden ball will serve the 
purpose. 



396 



BOOK VIII. SOLID GEOMETRY 



Proposition XII. Theorem 

667. In two polar triangles each angle of the one is 
the supplement of the opposite side in the other. 





Given two polar triangles ABC and A'B'C', the letter at the 
vertex of each angle denoting its value in degrees, and the small 
letter denoting the value of the opposite side in degrees. 

To prove that A + a' -180°, B + 5' = 180°, C + e' = 180°; 

A T +a =180°, B'+b =180°, C'+c =180°. 

Proof. Produce the arcs AB, AC until they meet B'C at 
the points D, E respectively. 

Since B' is the pole of AE, .'. B'E = 90°. 

And since C is the pole of AD, .'. DC' = 90°. 

.'. B'E + DC =180°. 

B'D + DE + DC = 180°, 

DE+ B'C = 180°. 

But DE is the measure of the Z. A, 

B'C = a'. 

.'. A+ a' =180°. 

B + b' = 180°, 

C + c' = 180°. 

In a similar way, starting with A A' B'C and producing the 
sides of A ABC, all the other relations are proved. q.e.d. 



That is, 
or 

and 

Similarly 
and 



§639 

Ax. 1 
Ax. 9 
Ax. 9 
§654 



SPHERICAL POLYGONS 



397 



Proposition XIII. Theorem 

668. The sum of the angles of a spherical triangle is 
greater than l$0° ami less than 540°, 





Given a spherical triangle ABC, the letter at the vertex of each 
angle denoting its value in degrees, and the small letter denoting 
the value of the opposite side in degrees. 

To prove that A + B + O180* and < 540°. 
Proof. Let A A'B'C " be the polar triangle of A ABC. 
Then A + a' = ISO . B + // = 180°, C + c' = 180°. § 667 

\\A+B + C + a' + V + c' = 540°. Ax. 1 

.-. a A- B + C = 540° - (V + /,' + c'). Ax. 2 

Now a' + b'A-c'<360 o . § 664 

.'. -1 + 2* + C = 540° — some value less than 360°. 
.-. A +5 + 0180°. 
Again a' + b' + c' is greater than 0°. 

.-. A+B+ C<540°. q.e.d. 

669. Corollary. A spherical triangle may have two, or 
even three, right angles; and a spherical triangle may havt 
two, or even three, obtuse angh s. 

670. Triangles classified as to Right Angles. A spherical 
triangle having two right angles is said to be birectangular; 
one having three right angli said to be trirectangular. 

The same terms may be applied to the corresponding trihedral angles. 



398 BOOK VIII. SOLID GEOMETKY 

EXERCISE 103 

1. If two sides of a spherical triangle are quadrants, the 
third side measures the opposite angle. 

2. In a birectangular spherical triangle the sides opposite 
the right angles are quadrants, and the side opposite the third 
angle measures that angle. 

Since the A are rt. A, what two planes are _L to a third plane ? What 
two arcs must therefore (§ 632) pass through the pole of a third arc ? 
Then what two arcs are quadrants ? Then how is the third angle (§ 654) 
measured ? 

3. Each side of a trirectangular spherical triangle is a 
quadrant. 

4. Three planes passed through the center of 
a sphere, each perpendicular to the other two, c( 
divide the spherical surface into eight congruent 
trirectangular triangles. 

Find the number of degrees in the sides of a spherical tri- 
angle, given the angles of its polar triangle as follows : 

5. 82°, 77°, 69°. 8. 83° 40', 48° 57', 103° 43'. 

6. 84i°, 81 j°, 72i°. 9. 96° 37' 40", 82° 29' 30", 68° 47'. 

7. 78° 30', 89°, 102°. 10. 43° 29' 37", 98° 22' 53", 87° 36' 39". 

Find the number of degrees in the angles of a spherical tri- 
angle, given the sides of its polar triangle as follows : 

11. 68° 42' 39", 93° 48' 7", 89° 38' 14". 

12. 78° 47' 29", 106° 36' 42", a quadrant. 

13. A quadrant, half a quadrant, three fourths of a quadrant. 

14. Erom the center of a sphere are drawn three radii, each 
perpendicular to the other two. Find the number of degrees 
in the sides and angles of the spherical triangle determined 
by their extremities. 




SPHERICAL POLYGONS 



399 




671. Symmetric Spherical Triangles. If through the center O 
of a sphere three diameters -4^4', BB', CC are drawn, and the 
points A,Bj C are joined by arcs of -rear circles, and also the 
]x>ints .1 './;'.' '. the two spherical tri- 
angles ABC and A'B'C'siTe called sym- 
metric spherical triangles. 

In the same way we may Eorm two sym- 
metric polygons of any number of sidi . 
Having thus formed the symmetric polygons, 
we may place them in any position we cho 
upon the surface of the sphere. 

672. Relation of Symmetric Triangles. Two symmetric tri- 
angles are mutually equilateral and mutually equiangular: yel 
in general they are not congruent, for the}' cannot be made to 
coincide by superposition. If in the above figure the triangle 
ABC is made to slide on the surface of the sphere until the 

rtex A falls on A', it is evident that the two triangles cannot 

be made to coincide for the reason that the corresponding parts 

of the triangles occur in reverse ordi 

To try to make two symmetric spherical polygons coincide is very 
much like trying to put the right-hand glove on the left hand. The rela- 
tion of two symmetric spherical triangles may be illustrated by cutting 
them out of the peel of an orange or an apple. 

673. Symmetric Isosceles Triangles. If, however, we have two 
symmetric triangles ABC and A'B'C' such that AB = ACj and 
A'B' = A'C' } that is, if the two sym- 
metric triangles are isosceles i then 
because AB, AC } A'B', A'C are all 
equal and the angles A and A' are 
equal, being originally formed by 
vertical dihedral angles i § 671), the B ir 
two triangles can be made to coincide Therefore, 

If two symmetric spherical triangles are isosceles^ they "re 
superpasable and tft congruent. 




400 BOOK VIII. SOLID GEOMETRY 

Proposition XIV. Theorem 
674. Tioo symmetric spherical triangles are equivalent. 




Given two symmetric spherical triangles ABC, A'B'C y having 
their corresponding vertices opposite each to each with respect to 
the center of the sphere. 

To prove that the triangles ABC, A'B'C are equivalent. 

Proof. Let P be the pole of a small circle passing through 
the points A, B, C, and let POP' be a diameter. 

Draw the great-circle arcsP.4, PB, PC, P'A', P'B', P'C'. 

Then PA=PB = PC. §636 

Now P'A' = PA, P'B' = PB, P'C = PC. § 672 

.'. P'A' = P'B' = P'C. Ax. 8 

.'. the two symmetric A PC A and P'C A' are isosceles. 

.'. A PC A is congruent to A P'C 'A'. § 673 

Similarly . A PAB is congruent to A P'A'B', 

and A PBC is congruent to A P'B'C. 

Now AABC = A PC A + A PAB -f A PBC, 

and A A 'B'C = A P'C A' + A P'A'B' + A P'B'C. Ax. 11 

.'.A ABC is equivalent to A A'B'C, by Ax. 9. q.e.d. 

Discussion. If the pole P should fall without the A ABC, then V 
would fall without A A'B'C, and each triangle would be equivalent to 
the sum of two symmetric isosceles triangles diminished by the third ; so 
that the result would be the same as before. 



SPHERICAL POLYGONS 



401 



Proposition XV. Theorem 

675. Two triangles on the sarru sphere or on equal 
spheres an either congruent or symmetric if two sides 
and the included angle of the one <<r< j respectively equal 
to tin corresponding parts of the other. 




Given two spherical triangles ABC and A'B'C, with AB=A 'B', 
AC = A'C, and angle A = angle A' , and similarly arranged ; and 
given the triangle A'B'X symmetric with respect to the triangle 
A'B'C. 

To prove that A ABC is congruent to A A'B'C, and that 
A ABC is symmetric with respect to A A'B'X. 

Proof. Superpose A .\J;<~ on A A'B'C, the proof being sim- 
ilar to that of the corresponding i-n+i' in plane geometry. § 68 

.-. A ABC is congruenl to A A'B'C. §662 

Since A A'B'X is symmetric with respect to A.! 'B'( '', 
and A ABC is congruent toA.-i7;V. 

.'. AC = A'X, AB = A'B', Z .1 = Z XA'B'. 

But A ABC is congruent to A A'B'C 1 and may be made to 
coincide with it. 

.*. A A BC is symmetric with respect to A A'B'X. q.e.d. 

Discussion. In the case of plane triangles, if the corresponding parts 
are arranged in reverse order, we can still prove the triangles congruent. 
Why can we no1 '1" so in the case of spherical triangli - 



402 



BOOK VIII. SOLID GEOMETRY 



Proposition XVI. Theorem 

676. Two triangles on the same sphere or on equal 
spheres are either congruent or symmetric if two angles 
and the included side of the one are respectively equal 
to the corresponding parts of the other. 




Given two spherical triangles ABC and A'B'C, with angle 
A = angle A', angle C— angle C, and AC = A'C, and similarly 
arranged ; and given the triangle A'B'X symmetric with respect to 
the triangle A'B'C. 

To prove that A ABC is congruent to A A'B'C, and that 
A ABC is symmetric ivith respect to A A'B'X. 

Proof. Superpose A ABC on A A'B'C', the proof being simi- 
lar to that of the corresponding case in plane geometry. 



§ i 



A ABC is congruent to A A'B'C 



inirii 



§662 



Since A A'B'X is symmetric with respect to A A'B'C', and 
A ABC is congruent to A A'B'C, 

.-. Z A = Z XA'B', AC = Z X, and AC = A'X. 

But Ayl.BC is congruent to A A'B'C" and may be made to 
coincide with it. 

.".A ABC is symmetric with respect to A A'B'X. q.e.d. 

Discussion. Under what circumstances are the two triangles both con- 
gruent and symmetric ? 

In plane geometry what is the case that corresponds to the one in 
which the spherical triangles are both congruent and symmetric ? 



SPHK I ! 1 1 ' -VL POLYGON S 



403 



Proposition XVII. Theorem 

677. Two mutually equilateral triangles on the same 
sphere or on equal spheres an mutually equiangular, 
and arc either congruent or symmetric. 




Given two spherical triangles, ABC, A'B'C, on equal spheres, 
such that AB = A'B', BC = B'C, CA = C'A'. 

To prove that AA = A A', AB = A B', Z C = Z C, and that 
/kABC and A'B'C are either congruent or symmetric 

Proof. Let and 0' be the centers of the spheres. 

Pass a plane through each pair of vertices of each triangle 

and the center of its sphere. 

Then in the trihedral angles at and 0' the face angles are 

equal each to its corresponding face angle. § 167 

.". the corresponding dihedral A are respectively equal. § 4 '.»'.» 
.*. the A of the spherical A are respectively equal. ?; <'>.V> 

.'. the A are either congruent or symmetric, by § , ''7<">. q.e.d. 

Discussion. In the figures the parts are arranged in the same order, 
so that the triangles are congruent. They might be arranged as in the 
figures of § 676. 

I>iscuss the proposition when the triangles are equilateral and each 
side is a quadrant. 

Discuss the proposition when two sides of each triangle air quadrants. 

What is the corresponding proposition in plane geometry, and why 
does not the form of proof there given hold here ? 



404 



BOOK VIII. SOLID GEOMETRY 



Proposition XVIII. Theorem 

678. Two mutually equiangular triangles on the same 
spliere or on equal spheres are mutually equilateral, 
and are either congruent or symmetric. 





Given two mutually equiangular spherical triangles T and T' on 
equal spheres. 

To prove that T and T are mutually equilateral, and are 
either congruent or symmetric. 

Proof. Let the AP be the polar triangle of A T, and the AP r 
be the polar triangle of A T'. 

Since the A T and T' are mutually equiangular, Given 

. * . the polar A P and P' are mutually equilateral. § 667 

.". the polar AP and P' are mutually equiangular. § 677 

But .the A T and T' are the polar A of A Pand P'. § 666 

.". the A T and T' are mutually equilateral. § 667 

Therefore the A T and V are either congruent or symmetric, 

by §677. Q.e.d. 

Discussion. The statement that mutually equiangular spherical tri- 
angles are mutually equilateral, and are either congruent or symmetric, 
is true only when they are on the same sphere or on equal spheres. When 
the spheres are unequal, the spherical triangles are unequal. In this case, 
however, their sides have the same arc measure, and therefore have the 
same ratio as the circumferences or as the radii of the spheres (§ 382). 



SPHERICAL POLYGONS 405 

Proposition XIX. Theorem 

679. In an isosceles spherical triangle the angles op- 
posite the equal sides are equal. 




Given the spherical triangle ABC, with AB equal to AC. 
To prove that ZB = ZC. 

Proof. Draw the arc AD of a great circle, from the vertex A 
to the mid-point of the base BC. 

Then AABD and ACD are mutually equilateral. 

.'. A ABD and A CD are mutually equiangular. § 677 

.'.AB = /-C. Q.e.d. 

EXERCISE 104 

1. The radius of a sphere is 4 in. From any point on the sur- 
face as a pole a circle is described upon the sphere with an open- 
ing of the compasses equal to 3 in. Find the area of this circle. 

2. The edge of a regular tetrahedron is </. Find the radii 
r, r' of the inscribed and circumscribed spheres. 

3. Find the diameter of the section of a sphere of diameter 
10 in. made by a plane 3 in. from the center. 

4. The arc of a great circle drawn from the vertex of an 
isosceles spherical triangle to the mid-point of the base l>is»- 
the vertical angle, is perpendicular to the base, and divides the 
triangle into two svmmetric triangles. 



406 



BOOK VIII. SOLID GEOMETRY 



Proposition XX. Theorem 

680. If two angles of a spherical triangle are equal, 
the sides opposite these angles are equal and the tri- 
angle is isosceles. 





Given the spherical triangle ABC, with angle B equal to angle C. 

To prove that AC ' = AB. 

Proof. Let A A'B'C be the polar triangle of A ABC. 

Since ZB = ZC, .-. A'C = A'B'. . §667 

.■.ZB' = ZC. §679 

.'.AC = AB j by § 667. Q.e.d. 

EXERCISE 105 

1. To bisect a given spherical angle. 

2. To construct a spherical triangle, given two sides and the 
included angle. 

3. To construct a spherical triangle, given two angles and the 
included side. 

4. To construct a spherical triangle, given the three sides. 

5. To construct a spherical triangle, given the three angles. 

6. To pass a plane tangent to a given sphere at a given point 
on the surface of the sphere. 

7. To pass a plane tangent to a given sphere through a given 
straight line without the sphere. 



SPHERICAL POLYGONS 407 

Proposition XXI. Theorem 

681. If two angles of a sph rical triangle are urn qual, 
the sides i>j>i><>.<<'it< ihesi angles are unequal, and the sidi 
opposite the greater angh is the greaii r; and if two sides 
(//■>■ unequal, the angles opposite these sides are unequal, 
and the angl opposite th greater side is the greater. 





Given the triangle ABC, with angle C greater than angle B. 

To prove that AB > AC. 

Proof. Draw the arc CD of a great circle, making A D CB 
equal to A B. Then DB = DC. § G80 

Now AD + DC> AC. §663 

.'. AD+ DB>AC, or AB>AC, by Ax. 9. q.e.d. 

Given the triangle ABC, with AB greater than AC. 

To prove that Z C is greater than Z /;. 

Proof. The Z C must be equal to, less than, or greater than 
the Z /;. 

If AC = AB, th.n AB = AC; § 680 

and if Z C is less than Z /?, then AB<A( \ as above. 
But both of these conclusions arc contrary to what is given. 
.'. AC is greater than All. q.e.d. 



408 BOOK VIII. SOLID GEOMETRY 

Proposition XXII. Theorem 

G82. The shortest line that can he drawn on the sur- 
face of a sphere between two points is the arc of a great 
circle joining the two points, not greater than a semi- 
circle. 



Given AB, the arc of a great circle, not greater than a semicircle, 
joining the points A and B. 

To prove that AB is the shortest line that can he drawn on 
the surface joining A and B. 

Proof. Let C be any point in A B. 

With A and B as poles and A C and BC as polar distances, 
describe two arcs DCF and GCE. 

The arcs DCF and GCE have only the point C in common. 
For if F is any other point in DCF, and if arcs of great circles 
AF and BF are drawn, then 

AF=AC. §636 

But AF+BF>AC + BC. §663 

Take away AF from the left member of the inequality, and 
its equal A C from the right member. 

Then BF>BC. Ax. 6 

Therefore BF>BG, the equal of BC. Ax. 9 

Hence F lies outside the circle whose pole is B, and the 
arcs DCF and GCE have only the point C in common. 



SPHERICAL POLYGONS 409 

Now let A DEB be any line from A to B on the surface of 
the sphere, which does not pass through C. 

This line will cut the arcs DCF and GCE in separate points 
D and E\ and if we revolve the line AD about A as a iixed 
point until D coincides with C, we shall have a line from A to 
C equal to the line A D. 

In like manner, we can draw a line from B to C equal to 
the line BE. 

Therefore a line can be drawn from A to B through C that 
is equal to the sum of the lines AD and BE, and hence is less 
than the line A DEB by the line DE. 

Therefore no line which does not pass through C can be the 
shortest line from A to B. 

Therefore the shortest line from A to B passes through C. 

But C is any point in the arc A B. 

Therefore the shortest line from A to B passes through 
every point of the arc AB, and consequently coincides with 
the arc AB. 

Therefore the shortest line from A to B is the great-circle 
arc A B. q.e.d. 

EXERCISE 106 

1. The three medians of a spherical triangle are concurrent. 

2. To construct with a given radius a spherical surface that 
passes through three given points. 

3. To construct with a given radius a spherical surface that 
passes through two given points and is tangent to a given plane. 

4. To construct with a given radius a spherical surface that 
passes through two given points and is tangent to a given 
sphere. 

5. The smallest circle on a given sphere whose plane passes 
through a given point within the sphere is the circle whose 
plane is perpendicular to the radius through the given point. 



410 



BOOK VIII. SOLID GEOMETRY 




683. Zone. A portion of a spherical surface included be- 
tween two parallel planes is called a zone. 

Thus on the earth we have the torrid zone included between the 
planes of the tropics of Cancer and Capricorn. 

The circles made by the planes are called the 
bases of the zone, and the distance between the 
planes is called the altitude of the zone. 

If one of the planes is tangent to the sphere and 
the other plane cuts the sphere, the zone is called a 
zone of one base. 

If both planes are tangent to the sphere, the zone 
is a complete spherical surface. 

684. Generation of a Zone. If a great circle revolves about 
its diameter as an axis, any arc of the circle generates a zone. 

Thus, in the figure of § 683, if the great circle PACQ revolves about 
its diameter PQ as an axis, the arc AC generates the zone A J), of which 
the altitude is the distance between the parallel planes. Similarly, the 
arc AP generates the zone ABP, and the arc CQ generates the zone CDQ, 
these both being zones of one base. 

685. Lune. A portion of a spherical surface bounded by 
the halves of two great circles is called a lune. 

Thus PAQB is a lune. A lune is 
evidently generated by the partial or 
complete revolution of half of a great 
circle about its diameter as an axis. 

686. Angle of a Lune. The angle 
between the semicircles bounding 
a lune is called the angle of the 
lune. 

Thus Z APB is the angle of the 
lune PA QB. 

A lune is usually taken as having an angle less than a straight angle. 
This is not necessary, for we may consider a hemispherical surface as a 
lune with an angle of 180°. We may also conceive of lunes with angles 
greater than a straight angle, and we may even think of an entire 
spherical surface as a lune whose angle is 360°. 




MEASUREMENT OF SPHERICAL SURFACES 411 



Proposition XXIII. Theorem 

687. The area of the surface generated by a straight 
Urn n wiving about an axis in its plane is equal to tin 
product of the projection of the line on the axis by 
tin circle ivhose radius is a perpendicular erected at tin 
midrpomt of the line ami terminated by the axis. 

A 31 



X I C O "TZ)"*" 




1 \ 


■-I b. 
i 


1 1 \ 

-I L_l. 

C OR 


i 

__L_ 

D 




Given an axis XY about which a line AB in the same plane with 
XY revolves, M being the mid-point of AB, CD being the projec- 
tion of AB on XY, MO being perpendicular to XY, MR being per- 
pendicular to AB, and a being the area generated by AB. 

To prove that a = CD x 2 ttMR. 

Proof. 1. If AB is II to IF, CD = AB, MR coincides with 
MO, and a is the lateral area of a right cylinder. § 588 

2. If AB is not II to A'}', and does not cut AT, a is the 
lateral area of the frustum of a cone of revolution. 

.-. a = AB x 2 it MO. 

Draw A E II to XY. 

The A AEB and MOR are similar. 

.-. MO: AE = MR : AB. 

.'. AB X MO = AE X MR, 

or AB X MO = CD x .1//,'. 

Substituting, a = CD x 2 77.1//,'. 

3. If A lies in the axis XY, then AE and CD coincide, 
and a = CD x 2ttMB, by § 609. q.e.d. 



§616 

§290 
§ 282 
§261 
Ax. 9 



412 



BOOK VIII. SOLID GEOMETRY 



Proposition XXIV. Theorem 

688. The area of the surface of a sphere is equal to 
the product of the diameter by the circumference of a 
great circle. 




E D' 



B' a 



Given a sphere generated by the semicircle ABCDE revolving 
about the diameter AE as an axis, 5 being the area of the surface, 
r being the radius, and d being the diameter. 

To prove that s=2 irrd. 

Proof. Inscribe in the semicircle half of a regular polygon 
having an even number of sides, as ABCDE. 

From the center O draw _k to the chords AB, BC, CD, DE. 
These J§ bisect the chords (§174) and are equal. § 178 

Let I denote the length of each of these _k 
From B, C, and D drop perpendiculars to AE. 
Then area of surface generated by AB = AB' x 2 nrl, § 687 

area of surface generated by BC = B'O x 2 ttI, etc. 
.*. area of surface generated by ABCDE = AE x 2 irl Ax. 1 

= 2 irld. Ax. 9 

Denote the area of the surface generated by ABCDE by s', 
and let the number of sides of ABCDE be indefinitely increased. 
Then s' approaches s as a limit, 

/ approaches r as a limit, 
and consequently 2 irld approaches 2 irrd as^ limit. 
But s' = 2 irld, always. 

• '.s = 2Trrd, by §207. 



§377 



§687 

Q.E.D. 



MEASUREMENT OF SPHERICAL SURFACES 413 

689. Corollary 1. The area of the surface of " sphere is 
equivalent to the area of four great circles, or to 4 irr 1 . 

In s = 2-rrrd, what is the value of d in terms of r ? Then what is the 
value of s in terms of r? 

Fur example, if the radius is 10 in., the area of the surface of the 
sphere is 4 7T- 100 sq. in., or 1266.64 sq. in. 

690. ( Orollary 2. The areas of the surfaces of two spheres 
are to each other as the squares on their radii, or as the squares 
on their diameters. 

If the radii are r and r', the diameters d and d', and the surfa 
s and $', then what is the ratio of .s to s', according to § 680 ? Show that 
this also equals r 2 : r' 2 , and d- : d" 1 . 

691. Corollary 3. The area of a zone is equal to the 
product of the altitude by the circumference of a great circle. 

If we apply the reasoning of § G88 to the zone generated by the revo- 
lution of the arc BCD, we obtain 

the area of zone BCD = B'T/ x 2 7rr, 

where B'T/ is the altitude of the zone and 2 7n- the circumference of 
a great circle. 

For example, if the radius is 10 in., and the altitude is 5 in., the area 
of the zone is 5 • 27r • 10 sq. in., or 314.16 sq. in. 

692. Corollary 4. The area of a zone of one base is equiv- 
alent to the area of a circle whose radius is the chord of th>- 
generating arc. 

The arc AB generates a zone of one base. 
.-. the area of the zone AB = AB' x 2 -nr = ttAB' x AE. 
But AB' x AE = AB 2 . § 298 

.-. the area of the zone AB = irAB". 

693. Spherical Excess of a Triangle. The excess of the sum of 
the angles of a spherical triangle over 180° is called the spin rim! 
excess of the triangle. 

For example, if the angles of a spherical triangle are 80°, 90°, and 100°, 
the spherical excess of the triangle is 90°. 



414 



BOOK VIII. SOLID GEOMETRY 



Proposition XXV. Theorem 

694. The area of a lime is to the area of the surface of 
the sphere as the angle of the lune is to four right angles. 





Given a lune PAQB, the great circle ABCD whose pole is P, a 
the value in degrees of the angle of the lune, / the area of the lune, 
and s the area of the surface of the sphere. 

To prove that I : s = a : 4 rt. A. 

Proof. The arc AB measures the Z a of the lune. § 654 

Hence arc AB : circle ABCD = a : 4 rt. A. § 212 

If AB and ABCD are commensurable, let their common meas- 
ure be contained m times in AB, and n times in ABCD. 

Then arc AB : circle ABCD = m : n. 

. ' . a : 4 rt. A = m : n. 

Pass an arc of a great circle through the poles P and Q and 
each point of division of ABCD. 

These arcs will divide the entire surface into n equal lunes, 
of which the lune PA QB will contain m. 

.'. 1: s = m : n. 

. • . I : s = a : 4 rt. A. Ax. 8 

If AB and ABCD are incommensurable, the theorem can be 
proved by the method of limits as in § 472. q.e.d. 



MEASUREMENT OF SPHERICAL SURFACES 415 

EXERCISE 107 

T~*hiiy it = 3.1416 for all examples in this exercise, find the 
areas of spheres whose radii are as follows: 

1. 2 in. 3. 3J in. 5. 2 ft. 1 in. 7. 48.8 in. 

2. 7 in. 4. 5§ in. 6. 3 ft. G in. 8. 4000 mi. 

Find tlic radii of spheres whose areas are a* follows: 

9. 12.5664 sq. in. 11. 1 sq. ft. 13. s. 

10. 50.2656 sq. in. 12. 100 tt sq. in. 14. 4 7T 3 . 

On a sphere whose radius is 20 in., find the areas of zones 
whoSi <dfitudes are as folloivs : 

15. 2 in. 17. 7 in. 19. 1 ft. 21. 3.45 in. 

16. 3 in. 18. 10 in. 20. 2} 2 in. 22. 6.83 in. 

On a sphere whose radius is 10 in., find the areas of lanes 
whose angles are as folloivs : 

23. 30°. 25. 90°. 27. 22° 30'. 29. 52° 20' 20". 

24. 45°. 26. 180°. 28. 7° 30'. 30. 48° 35' 10". 

31. Two limes on the same sphere or on equal spheres have 
the same ratio as their angles. 

32. The area of a lune is equal to one ninetieth of the area 
of a great circle multiplied by the number of degrees in the 
angle of the lune. 

33. Zones on the same sphere or on equal spheres are to 
each other as their altitudes. 

34. Given the radius of a sphere 15 in., find the area of a 
lune whose angle is 30°. 

35. Given the diameter of a sphere 16 in., find the area <»f a 
lune whose angle is 75°. 

36. What is the spherical excess of a trirectangular triangle ? 



416 



BOOK VIII. SOLID GEOMETRY 



Proposition XXVI. Theorem 

695. A spherical triangle is equivalent to a lune whose 
angle is half the spherical excess of the triangle. 





Given the spherical triangle ABC on a sphere of surface s. 

To prove that A ABC is equivalent to a lune ivhose angle is 
\(/.A + Z.B + Z.C- 180°). 

Proof. Produce the sides of the A ABC to complete circles. 
Now AAB'C and A'BC are symmetric. Const. 

.*. A AB'C is equivalent to A A'BC. § 674 

.'.lune ABA'C = A ABC + AAB'C. Ax. 9 



But 



ACB'A +AAC'B + AAB'C' + AABC = bs. Ax. 11 
.'. (lune BCB'A — A ABC) + (lune CAC'B — A ABC) 

4- lune ABA'C = J s. Ax. 9 

.-. 2 A ABC = lune BCB'A -f- lune CAC'B 

+ lune ABA 'C — i s. Axs. 1, 2 

.-.A 4£C = i (lune J5C5U + lune CMC'£ 

+ hme ABA'C — % s). 
But i s = a lune whose angle is 180°. 

.'. A ABC = a lune whose angle is 

\(/-A +ZB + ZC -ISO ). q.e.d. 

Discussion. Since we have found (§ 694) how to compute the area of 
a lune, we can now compute the area of a spherical triangle when the 
angles are known. 



Ax. 4 
§694 



MEASUREMENT OK SPHERICAL SURFACES 417 

696. Corolla kv. //" tiro great-circle arcs intersect within 
a great circle, the sum of the two opposite spherical triangl 8 
which they form with the great circle is ,>piienlent to a lime 
whose angle is the angle between the arcs. 

697. Computation of Area. To illustrate the computation in- 
volved in §695, findthearea of a triangle whose angles are 110°, 
100°, and 95°, on the surface of a sphere whose radius is G in. 

Spherical excess = 110° + 100° + 95° - 180 3 = 125°. 

.-. angle of lime = <;•_>' . 

02 1 
.-. area of lime = — - of the spherical surface. 
360 

.-. area of lime = — - x 4 x 3.1416 x 36 sq. in. 
360 

.-. area of triangle = 78.54 sq. in. 

698. Spherical Excess of a Polygon. The excess of the sum of 
the angles of a spherical polygon of n sides over (n — 2) x 180° 
is called the spherical excess of the polygon. 

EXERCISE 108 

Compute the areas of triangles on spheres of the given diam- 
eters, the angles being as follows: 

1. 100°, 120°, 140°, il =16 in. 4. 115°, 124°, 85°, d = 30 in. 

2. 105°, 130°, 125°, d =10 in. 5. 135°, 110°, 92°, d = 40 in. 

3. 127°, 132°, 90°, d = 20 in. 6. 148°, 93°, 68°, d = 25.8 in. 

7. 115° 27' 30", 102° 32' 48", 08° 27 ; 39", d = 8000 mi. 

Compute the areas of triangles on spheres of the given radii, 
the angles being as follows: 

8. 120°, 100°, 90°, r = 9 in. 11. 115°, 102°, 30°, r = 36 in. 

9. 130°, 90°, 80°, r=10 in. 12. 1 10°, 120°, 85°, r = 90in. 
(10. 105°, 75°, (35°, r = 18 in. 13. 130°, 117°, 93°, r = 1.8 in. 



418 BOOK VIII. SOLID GEOMETRY' 

Compute the areas of triangles on spheres of the given cir- 
cumferences, the angles being as folloivs : 

14. 93°, 94°, 120°, c = 31.416 in. 

15. 82°, 105°, 98°, c = 62.832 in. 

16. 148°, 27°, 125°, c = 15.708 in. 

17. 162°, 39°, 120°, c = 78.54 in. 

18. 149°, 41°, 116°, c = 39.27 in. 

19. 126° 30' 42", 105° 26' 15", 63° 15' 3", c = 314.16 in. 

20. What is the area of a triangle on the earth's surface 
the vertices of which are the north pole and two points on the 
equator, one at 37° W. and the other at 16° E., the earth being 
considered a sphere with a radius of 4000 mi. ? 

21. If the radii of two spheres are 6 in. and 4 in. respec- 
tively, and the distance between the centers is 5 in., what is 
the area of the circle of intersection of the spheres ? 

22. Find the radius of the circle determined on a sphere of 
5 in. diameter by a plane 1 in. from the center. 

23. If the radii of two concentric spheres are r and r', and 
if a plane is passed tangent to the interior sphere, what is the 
area of the section made in the other sphere ? 

24. Two points A and B are 8 in. apart. Find the locus in 
space of a point 5 in. from A and 7 in. from B. 

25. Two points A and B are 10 in. apart. Find the locus in 
space of a point 7 in. from A and 3 in. from B. 

26. The radii of two parallel sections of the same sphere are 
a and b respectively, and the distance between the sections is 
d. Find the radius of the sphere. 

27. The diameter of a certain sphere is V2. The chords of 
the arcs that form the sides of a triangle on the surface of the 
sphere are respectively 1,1, and £ V2. Find the area of the 
spherical triangle. 



MEASUREMENT OF SPHERICAL SURFACES 419 

Proposition XXVII. Theorem 

699. A spin rical polygon is > quivalent to a lune ivhose 
angle is half tin spJierical excess of the polygon. 




Given a spherical polygon P of n sides, the sum of the angles 
being s. 

To prove that P is equivalent to a lune whose angle is 
^( s _^T^2xl80°). 

Proof. Draw all the diagonals from any vertex. 

Since there is a distinct triangle for each side except those 
meeting at the vertex chosen, there are (n — 2) triangles. 

Since each triangle is equivalent to a lune whose angle is 
half the excess of the sum of its angles over 180°, § 695 

therefore the (?i — 2) triangles are equivalent to a lune whose 
angle is half the excess of the sum of all the angles of the 
polygon over (n — 2) X 180°. 

.*. P = a lune whose angle is J(s — n — 2 x ISO ). Q. B. d. 

700. Computation of Area. Find the area of a spherical poly- 
gon whose angles are 100°, 110°, 120°, and 170°, r bring 6 in. 
Spherical excess = 100° + 110° + 120° + 170° - 2 x 180° = 140°. 



-, i 



angle of lune =: 70 



area of lune = jfc of 4 irr' 2 

= ^ of 4 x 3.1410 x 30 sq. in. 
= 87.9048 sq. in. 



420 BOOK VIII. SOLID GEOMETRY 

EXERCISE 109 

Find the areas of spherical polygons on spheres of the given 
areas, the angles being as follows : 

1. 30°, 90°, 120°, 130°, a =2 sq. ft. 

2. 45°, 60°, 100°, 165°, a = 288 sq. in. 

3. 70°, 168°, 92°, 120°, a = 500 sq. in. 

4. 68° 30 r , 149° 50', 96° 54', 136° 52', a = 750 sq. in. 

5. 122°27'40", 130°32'50", 98°31'30", 96° 48', a = 600sq.in. 

6. 132°, 96°, 154°, 120°, 150°, a = 3 sq. ft. 120 sq. in. 

7. 130°, 156°, 172°, 95°, 120°, 100°, a = 157.2 sq. in. 

Find the areas of spherical polygons on spheres of the given 
radii, the angles being as follows : 

8. 130°, 150°, 80°, 90°, r = 10 in. 

9. 148°, 157°, 90°, 100°, 120°, r = 20 in. 

10. 172°, 169°, 86°, 141°, 100°, 90°, r = 24 in. 

11. 135° 30', 148° 42', 96° 37', 102° 11', r = 10 in. 

Find the areas of spherical polygons on spheres of the given 
diameters, the angles being as follows: 

12. 148°, 92°, 60°, 120°, d = 10 in. 

13. 172°, 168°, 93°, 37°, 100°, ^ = 22 in. 

14. 102°, 162°, 139°, 141°, 138°, 126°, d = 20 in. 

15. 82°50'42", 120° 29' 18", 98°37'15", 141°22'45", d = 20 in. 

Find the areas of spherical polygons on spheres of the given 
circumferences, the angles being as follows : 

16. 39°, 148°, 172°, 168°, c = 3.1416 in. 

17. 128°, 92°, 168°, 109°, c = 31.416 in. 

18. 146°, 129°, 102°, 137°, 100°, c = 6.2832 in. 

19. 128°, 145°, 139°, 82°, 161°, 137°, c = 18.8496 in. 



MEASUREMENT OF SPHERICAL SOLIDS 421 



701. Spherical Pyramid. A portion of a sphere bounded by 
a spherical polygon and the planes of its 
sides is called a spherical pyramid. 

The center of the sphere is called the vert< x 
of the spherical pyramid, and the spherical 
polygon is called the base. 

Thus O-ABC is a spherical pyramid. 

702. Spherical Sector. A portion of a sphere generated by 
the revolution of a circular sector about any diameter of the 
circle of which the sector is a part is called a spherical sector. 






Thus if the sector A OB revolves about the diameter MX as an axis, it 
generates the spherical sector AB-O-A'B'. 

The zone generated by the arc of the generating sector is called the 
base of the spherical sector. 

703. Spherical Segment. A portion of a sphere contained 
between two parallel planes is called a spherical segment. 

The sections of the sphere made by the parallel planes are called the 
bases of the spherical segment, and the distance between these bases is 
.ailed the altitude of the spherical segment. 

If one of the parallel planes is tangent to the sphere, the segment 
is called a spherical segment of one base. 

A spherical segment of one base may be generated by the revolution 
of a circular segment about the diameter perpendicular to its bi 

704. Spherical Wedge. A portion of a sphere bounded by a 
lune and the planes of two great circles is called a spherical 
wedge. 



422 



BOOK VIII. SOLID GEOMETRY 



Proposition XXVIII. Theorem 

705. The volume of a sphere is equal to the product 
of the area of its surface by one third of its radius. 





Given a sphere of radius r, area of surface s, volume z;, and 
center 0. 



v = sx\r. 



To prove that 

Proof. We may imagine a cube of edge 2 r circumscribed 
about the sphere. 

Connect O with each of the vertices of this cube. 

These connecting lines are the edges of six pyramids whose 
bases are the faces of the cube and whose altitudes all equal r. 

The volume of each pyramid is a face of the cube multiplied 
by ^r, and the volume of the six pyramids, or of the whole 
cube, is the area of the surface of the cube multiplied by ^ r. 

Xow imagine planes drawn tangent to the sphere, at the 
points where the edges of the pyramids cut its surface. We 
then have a circumscribed solid whose volume is nearer that 
of the sphere than is the volume of the circumscribed cube, 
but is still greater than the sphere. Ax. 11 

Proceeding as before, connect with the vertices of the 
new polyhedron. These connecting lines are the edges of 
pyramids whose bases are together equal to the bases of the 
polyhedron and whose common altitude is r. § 646 



MEASUREMENT OF SPHERICAL SOLIDS 423 

Then the sum of the volumes of these pyramids is again the 
area of the surface of the polyhedron multiplied by £ r. De- 
noting this volume by v' and the area of the surface by s', we 

have v' = s' x J r. 

• 

If we continue to draw tangent planes to the sphere, we con- 
tinue to diminish the circumscribed solid. 

By continuing this process indefinitely we can make the 
difference between the volume of the sphere and the volume 
of the circumscribed solid less than any assigned positive 
quantity, however small, the difference between the surface of 
the sphere and the surface of the circumscribed solid becoming 
and remaining less than any assigned value, however small. 

.\ v is the limit of v', and s is the limit of s'. § 204 

And since it has been shown that 

v' = s r X% i', always, 

,\v = sx£ r, by § 207. q.e.d. 

706. < < .rollary 1. The volume of a sphere of radius r and 

diann ter d is equal to |- 7T/* 3 or 1 ird^. 

For in v = s x | r what is the value of s in terms of r ? What is the 
value of d in terms of r '? Then what is the value of v in terms of d ? 

707. Corollary 2. The volumes of two spheres are to each 

other as the cubes of their radii. 

What is the ratio of f 7TT 3 to f irr' z ? 

By the same reasoning, the volumes are to each other as the cubes of 
the diameters. 

708. Corollary 3. The volume of a spherical sector is 

equal to one third the product of the area of the zone which 

forms its base multiplied 1>ij the rad'mx of the sphere. 

Suppose the base divided into spherical triangles. The planes deter- 
mined by their vertices are the bases of triangular pyramids with ver- 
tices at O. What is the limit of the sum of the volumes of these pyramids 
as the bases decrease in size ? 



424 BOOK VIII. SOLID GEOMETRY 

EXERCISE 110 
Problems of Computation 
Find the volumes of spheres whose radii are : 

1. 3 in. 4. 2i in. 7. 20.7 ft. 

2. 5 in. 5. 4| in. 8. 2 ft. 3 in. 

3. 7 in. 6. 9$ in. 9. 4000 mi. 

Find the volumes of spheres whose diameters are :. 

10. 24 in. 13. 2.8 in. 16. 2 ft. 1 in. 

11. 36 in. 14. 3.4 in. 17. 3 ft. 4 in. 

12. 48 in. 15. 4.5 in. 18. 8 ft. 6 in. 

Find the volumes of spheres whose circumferences are : 
19. 6.2832 in. 20. 12.5664 in. 21. 18.8496 in. 

Find the volumes of spheres whose surface areas are : 

22. 12.5664 sq. in. 23. 50.2656 sq. in. 24. 113.0976 sq. in. 

Find the radii of spheres whose volumes are : 

. 25. 4.1888 cu. in. 26. 33.5104 cu. in. 27. 113.0976 cu. in. 

28. The circumference of a hemispherical dome is 66 ft. 
How many square feet of lead are required to cover it ? 

29. If the ball on the top of St. Paul's Cathedral in London 
is 6 ft. in diameter, how much would it cost to gild it at 9 cents 
per square inch ? 

30. The dihedral angles made by the faces of a spherical 
pyramid are 80°, 100°, 120°, and 150°, and the length of a 
lateral edge is 42 ft. Pind the area of the base. 

31. The dihedral angles made by the faces of a spherical 
pyramid are 60°, 80°, and 100°, and the area of the base is 
4 7r sq. ft. Pind the radius. 



EXERCISES 425 

32. What is the area of the surface of the earth ? 

Assume that the earth is a sphere with a radius of 4000 mi., and make 
the same assumption in subsequent examples relating to the earth. 

33. The altitude of the torrid zone is 3200 mi. Find its area. 

34. What is the area of the north temperate zone if its 
altitude is 1800 mi. ? 

35. Find the number of square miles of the earth's surface 
that can be seen from an aeroplane 1500 ft. above the surface. 

36. How far in one direction can a man see from the deck 
of an ocean steamer if his eye is 40 ft. above the water? 

37. To what height must a man be raised above the earth 
in order to see one sixth of its surface ? 

38. How much of the earth's surface w^ould a man see if he 
were raised to the height of the radius above it ? 

39. If the atmosphere extends 50 mi. above the surface of 
the earth, find the volume of the atmosphere. 

40. If an iron ball 4 in. in diameter weighs 9 lb., find the 
weight of a spherical iron shell 2 in. thick, the external diame- 
ter being 20 in. 

41. What is the angle of a spherical wedge if its volume is 
li cu. ft. and the volume of the entire sphere is 8?- cu. ft. ? 

42. The inside of a washbasin is in the shape of the segment 
of a sphere. The distance across the top is 16 in. and its 
greatest depth is 8 in. How many pints of water will it hold, 
allowing 7 gal." to the cubic foot ? 

43. Prove that the volume of a spherical pyramid is equal 
to the product of the base by one third of the radius, and find 
the volume if the base is one eighth of the surface of a sphere 
of radius 10 in. 

44. Find the volume of a spherical sector whose base is a 
zone of area .-.. the radius of the sphere being >■, following a 
process of reasoning similar to that in § 705. 



426 BOOK VIII. SOLID GEOMETKY 

EXERCISE 111 
Formulas 

1. Find the area z of the zone of a sphere of radius r, illu- 
minated by a lamp placed at the height h above the surface. 

2. Find the volume v of a sphere in terms of c, the cir- 
cumference. 

3. Find the radius r of a sphere in terms of v, the volume. 

4. Find the diameter d of a sphere in terms of s, the 
area of the surface. 

5. Find the circumference c of a sphere in terms of s, the 
area of the surface. 

6. What is the altitude a of a zone, if its area is z and the 
volume of the sphere is v ? 

7. Show that in a spherical pyramid v = ^ br. Find r in 
terms of v and b ; also b in terms of v and r. 

8. Find a formula for the volume of the metal in a spher- 
ical iron shell, the inside radius being r and the thickness of 
the metal being t. 

9. Find a formula for the weight of a spherical shell, the 
inside radius being r, the thickness of the metal being t, and 
the weight of a cubic unit of metal being w. 

10. If the area of a zone z equals 2 irra (§ 691), find a for- 
mula for a in terms of z and r. 

11. If the area of a zone is expressed by the formula z = 2 irra, 
what is the diameter of the sphere upon which a zone z has an 
altitude a ? 

12. Find the area z of a zone of altitude a on a sphere whose 
area of surface is 5. 

13. Find a formula for the area a of that part of the surface 
of a sphere of radius r seen from a point at a distance d above 
the surface. 



EXERCISES 427 

EXERCISE 112 
Problems of Loci 

Find the locus of a point : 

1. At a given distance from a given point. 

2. At a given distance from a given straight line. 

3. At a given distance from a given plane. 

4. At a given distance from a given cylindric surface. 

5. At a given distance from a given spherical surface. 

6. Equidistant from two given points. 

7. Equidistant from two given planes. 

8. At a given distance from a given point and at another 
given distance from a given straight line. 

9. At a given distance from a given point and at another 
iven distance from a given plane. 

10. At a given distance from a given point and equidistant 
from two other given points. 

11. At a given distance from a given point and equidistant 
from two given planes. 

Find one or more points : 

12. At a distance d 1 from a given point, at a distance d 2 from 
a given straight line, and at a distance </ 3 from a given plan.-. 

13. At a distance d x from a given point, at a distance d 2 
from a given plane, and equidistant from two other given 
planes. 

14. Equidistant from two given points, equidistant from two 
given planes, and at a distance /• from a given point. 

15. Find the locus of the center of a sphere whose Burface 
touches two given planes and passes through two given points 
that lie between the planes. 



ir 



428 BOOK VIII. SOLID GEOMETKY 

EXERCISE 113 
Miscellaneous Exercises 

1. The volume of a sphere is to the volume of the inscribed 
cube as ir is to § V 3. 

2. The volume of a sphere is to the volume of the circum- 
scribed cube as it is to 6. 

3. Find the ratio of the volume of a cube inscribed in a 
sphere to that of a circumscribed cube. 

4. Find the difference between the volumes of two cubes, 
one inscribed in a sphere of radius 10 in. and the other circum- 
scribed about it. 

5. The planes perpendicular to the three faces of a trihedral 
angle, and bisecting the face angles, meet in a straight line. 

6. The planes that pass through the edges of a trihedral 
angle, and are perpendicular to the opposite faces, meet in a 
straight line. 

7. The altitude of a regular tetrahedron is equal to the sum 
of four perpendiculars let fall from any point within the tetra- 
hedron upon the four faces. 

8. To cut a given tetrahedral angle by a plane so that the 
section shall be a parallelogram. 

9. Compare the volumes of the solids generated by the 
revolution of a rectangle successively about two adjacent sides, 
the sides being a and b respectively. 

10. Find the difference between the volume of a frustum of 
a pyramid and the volume of a prism each 24 ft. high, if the 
bases of the frustum are squares with sides 20 ft. and 16 ft. 
respectively, and the base of the prism is the section of the 
frustum parallel to the bases and midway between them. 

11. To draw a line through the vertex of any trihedral angle, 
making equal angles with its edges. 



EXERCISES 429 

12. The lines drawn from each vertex of a tetrahedron to 
the point of intersection of the medians of the opposite face all 
meet in a point called the center of gravity of the tetrahedron, 

which divides cadi line so that the ratio of the Bhorter seg- 
ment to the whole line is 1 : 4. 

13. The lines joining the mid-points of the opposite edges 
of a tetrahedron all pass throngh the enter of gravity and are 

ected by it. 

14. The plane which bisects a dihedral angle of a tetrahe- 
dron divides the opposite edge into segments proportional to 
the areas of the faces that include the dihedral angle. 

15. To cut a given cube by a plane so that the section shall 
be a regular hexagon. 

16. The volume of a right circular cylinder is equal to the 
product of the lateral area by half the radius. 

17. The volume of a right circular cylinder is equal to the 
product of the area of the rectangle which generates it, by the 
length of the circumference generated by the point of intersec- 
tion of the diagonals of the rectangle. 

18. If the altitude of a right circular cylinder is equal to 
the diameter of the base, the volume is equal to the total area 
multiplied by a third of the radius. 

19. The surface of a sphere is two thirds the total surface 
of the circumscribed cylinder. 

20. The volume of a sphere is two thirds the volume of the 
circumscribed cylinder. 

21. Given a sphere, a cylinder circumscribed about the 
sphere, and a cone of two nappes inscribed in the cylinder. 
If any two planes are drawn perpendicular to the axis of the 
three figures, the spherical segment between the planes is 
equivalent to the difference between the corresponding cylin- 
dric and conic segments. 



430 BOOK VIII. SOLID GEOMETRY 

EXERCISE 114 

Review Questions 

1. How is a sphere generated ? 

2. What are two tests of equality of spheres ? 

3. If a plane cuts a sphere, what figure is formed ? Is the 
same true of a plane cutting a cone ? 

4. What is the test of equal circles on a given sphere ? 

5. What is a great circle of a sphere ? Name four proper- 
ties of great circles. 

6. What is meant by a plane being tangent to a sphere ? 
State any proposition concerning a tangent plane, and the cor- 
responding proposition in plane geometry. 

7. Complete this statement: A sphere may be inscribed 
in • • . . State the corresponding proposition in plane geometry. 

8. Complete this statement : A sphere may be circum- 
scribed about • • •. State the corresponding proposition in 
plane geometry. 

9. Complete this statement: A spherical surface is deter- 
mined by • • • points not in the same plane. State the corre- 
sponding proposition in plane geometry. 

10. What is the limit of the sum of the sides of a spherical 
polygon ? What are the limits of the sum of the angles of a 
spherical triangle ? 

11. What is a polar triangle? State two propositions re- 
lating to polar triangles. 

12. What is meant by symmetric spherical triangles ? State 
two propositions relating to such triangles. 

13. State two propositions relating to congruent spherical 
triangles. 

14. How is the area of a spherical triangle found ? How is 
the area of a spherical polygon found ? 



APPENDIX 

709. Subjects Treated. As with plane geometry, bo with 
solid geometry, there are many topics that might be taken in 
addition to those given in any textbook. The theorems and 
problems already given in this work are standard propositions. 
that are looked upon as basal, and are usually required as 
preliminary to more advanced work, and these, with a reason- 
able selection from the exercises, will be all that most schools 
have time to consider. It occasionally happens, however, that 
a school is able to do more than this, and then more exercises 
may be selected from the large number contained in this work, 
and a few additional topics may be studied. For this latter 
purpose the appendix is added, but its study should not be 
undertaken at the expense of good work on the fundamental 
propositions and the exercises depending upon them. 

The subjects treated are certain additional propositions in 
the mensuration of solids, and a few general theorems relating 
to similar polyhedrons, these being occasionally required for 
college examinations. There is also added a brief sketch of the 
history of geometry, which all students are advised to read as 
a matter of general information, and a few of those recreations 
of geometry that add a peculiar interest to the subject. 

710. Similar Polyhedrons. Polyhedrons that have the saim- 
number of faces, respectively similar and similarly placed, and 
their corresponding polyhedral angles equal, are called similar 
polyhedrons. 

It will be seen that this is analogous to the definition of similar 
polygons in piano geometry. 

431 



432 



APPENDIX TO SOLID GEOMETRY 



Proposition I. Theorem 



711. A truncated triangular prism is equivalent to 
the sum of three pyramids ivhose common base is the 
base of the prism and ivhose vertices are the three ver- 
tices of the inclined section. 





Given a truncated triangular prism ABC-DEF whose base is 
ABC and inclined section DEF, the truncated prism being divided 
into the three pyramids E-ABC, E-ACD, and E-CFD. 

To prove ABC-DEF equivalent to the sum of the three pyr- 
amids E-ABC, D-ABC, and F-ABC. 

Proof. E-ABC has the base ABC and the vertex E. 

Now pyramid E-A CD = pyramid B-A CD. § 558 

{For they have the same base, ACD, and the same altitude, since their 
vertices E and B are in the line EB \\ to the plane ACD.) 

But the pyramid B-A CD may be regarded as having the base 
ABC and the vertex D ; that is, as pyramid D-ABC. 



PRISMS 



433 



Then since A CFD and ACF have the common base CFand 
equal altitudes, their vertices lying in the line A I) which is 
parallel to CF, they are equivalent. § 326 

Furthermore, pyramids E-CFD and B-ACF not only have 
equivalent bases, the A CF/> and A < ' F. hut they have the same 
altitude, since their vertices E and />' are in the line EB which 
is parallel to the plane of their bases. 

.'. pyramid E-CFD = pyramid B-ACF. § 558 

But the pyramid B-A CF may be regarded as having the base 
ABC and the vertex F; that is, as pyramid F-ABC. 

Therefore the truncated triangular prism ABC-DEF is 
equivalent to the sum of the three pyramids E-ABC, D-ABC, 
and F-ABC. q.e.d. 





712. Corollary 1. The volume of a truncated right tri- 
angular prism is equal to the product of its base by one third 
the sum of its lateral edges. 

For the lateral edges DA, EB, FC (Fig. 1), being perpendicular to 
the base ABC, are the altitudes of the three pyramids whose sum is 
equivalent to the truncated prism. It is interesting to consider the spe- 
cial case in which ADEF is parallel to A ABC. 

.713. Corollary 2. The volume of any truncated triangular 
prism is equal to the product of its right section by one third 
the sum of its lateral edges. 

For the right section DEF (Fig. 2) divides the truncated prism into 
two truncated right prisms. 



434 



APPENDIX TO SOLID GEOMETRY 



Proposition II. Theorem 

714. The volumes of two tetrahedrons that have a 
trihedral angle of the one equal to a trihedral angle 
of the other are to each other as the products of the 
three edges of these trihedral angles. 




Given the two tetrahedrons S-ABC and S'-A'B'C, having the 
trihedral angles S and S' equal, v and v' denoting the volumes. 

rr + -l , v SAxSBxSC 
To prove that — = — — - 

F v' S'A' x S'B' x S'C 

Proof. Place the tetrahedron S-ABC upon S'-A'B'C so that 
the trihedral Z S shall coincide with the equal trihedral Z S ! . 

Draw CD and CD' J_ to the plane S'A'B', 

and let their plane intersect S'A'B' in S'DD'. 

The faces S'AB and S'A'B' may be taken as the bases, and 
CD, C'D' as the altitudes, of the triangular pyramids C-S'AB 
and C-S'A'B' respectively. 



Then 
But 

and 

v 



1 S ' AB x CD 

v' ~ ' S'A'B' x CD' 
S'AB 
S'A'B' 
CD 



S'AB CD 
X 



S'A'B' " CD' 
S'A X S'B 
S'A' X S'B'' 
S'C 



CD' S'C 



? i n i 



§562 
§332 

§282 



S'A X S'B X S'C 



S'A' X S'B' X S'C S'A' X S'B' X S'C 



SAxSBxSC , . ft 

— ; j by Ax. 9. Q.E.D. 



POLYHEDRONS 



Proposition III. Theorem 



i:;;> 



715. In any polylinlron the number of edges increased 
by two is equal to the number of vertices increased by 
the number of faces. 




Given the polyhedron AG, e denoting the number of edges, v the 
number of vertices, and / the number of faces. 

To prove that e + 2 = v +/. 

Proof. Beginning with one face BCGF, we have e — v. 

Annex a second face A B CD by applying one of its edges to 
a corresponding edge of the first face, and there is formed a 
surface of two faces having one edge BC and two vertices B 
and C common to the two faces. 

Therefore for two faces e = v -f-1. 

Annex a third face ABFE, adjoining each of the first two 
faces. This face will have two edges AB, BF and three ver- 
tices A, B, F in common with the surface already formed. 

Therefore for three faces e = v + 2. 

In like manner, for four faces, e = c + 3, and so on. 

Therefore for (/- 1) faces e = v + (/- 2). 

But/— 1 is the number of faces of the polyhedron when 
only one face is lacking, and the addition of this face will not 
increase the number of edges or vertices. Hence for / faces 

e= v+f—2, ore + 2= v+f. q.e.d. 

This theorem is due to the gnat Swiss mathematician, Euler. 



436 APPENDIX TO SOLID GEOMETKY 

Proposition IV. Theorem 

716. The sum of the face angles of any polyhedron is 
equal to four right angles taken as many times, less 
two, as the polyhedron has vertices. 




Given the polyhedron P, e denoting the number of edges, v the 
number of vertices, / the number of faces, and 5 the sum of the 
face angles. 

To prove that s = (y — 2) 4 rt. A. 

Proof. Since e denotes the number of edges, 2 e will denote 
the number of sides of the faces, considered as independent 
polygons, for each edge is common to two polygons. 

If an exterior angle is formed at each vertex of every poly- 
gon, the sum of the interior and exterior angles at each vertex 
is 2 rt. A ; and since there are 2 e vertices, the sum of the 
interior and exterior angles of all the faces is 

2 e x 2 rt. A, or e x 4 rt. A. 

But the sum of the ext. A of each face is 4 rt. A. § 146 

Therefore the sum of all the ext. A of / faces is 

/ x 4 rt. A. 
Therefore s = e X 4 rt. A — f X 4 rt. A 

= (e _/) 4 rt. A. 

But e + 2 = v +/; ■ § 715 

that is, e— f=v — 2. Ax. 2 

Therefore s = (v — 2) 4 rt.. A. Q. e. d. 



POLYHEDRONS 437 

EXERCISE 115 

Find the volumes of truncated triangular prisms, given the 
fntsi's />, mill tin distances of the three vertices p, q, r from 
the planes of the bases, as follows: 

1. h = 8 sq. in., p = ."> in., q = 4 in., r — 5 in. 

2. 1) = ( .) sq. in., y> = (5 in., */ = 3 in., r = 4^ in. 

3. A = 15 sq. in.. /- == 7 in., y = 9 in., r = 8.1 in. 

4. 6 = 32 sq. in., y> = 9 in., <[ = VI in., v = 9.3 in. 

5. // = 48 sq. in., /> = 16 in., «/ = 15 in., r = 18 in. 

6. A triangular rod of iron is cut square off (i.e. in right 
section) at one end, and slanting at the othef end. The right 
section is an equilateral triangle li in. on a side. The edges of 
the rod are 3 ft. 2 in., 3 ft. 3 in., and 3 ft. 3 in. Find the weight 
of the rod, allowing 0.28 lb. per cubic inch. 

7. Two triangular pyramids with a trihedral angle of the 
one equal to a trihedral angle of the other have the edges of 
these angles 3 in., 4 in., 3^ in., and 5 in., b\ in., 6 in. respec- 
tively. Find the ratio of the volumes. 

8. Make a table giving the number of edges, vertices, and 
faces of each of the five regular polyhedrons, showing that in 
every case the number conforms to Euler's theorem (§ 715). 

9. Make a table similar to that of Ex. 8, giving the sum 
of the face angles in each of the five regular polyhedrons, 
showing that in every case s = (v — 2) 4 rt. Zs (§ 716). 

10. There can be no seven-edged polyhedron. 

11. Can there be a nine-edged polyhedron ? 

12. What is the sum of the face angles of a six-edged poly- 
hedron ? 

13. What is the sum of the face angles of a polyhedron 
with five vertices? with four vert ices ? Consider the possi- 
bility of a polyhedron with three vertices. 



438 



APPENDIX TO SOLID GEOMETEY 



Proposition V. Theorem 

717. Two similar polyhedrons can be separated into 
the same number of tetrahedrons similar each to each 
and similarly placed. 




Given two similar polyhedrons P and P 1 . 

To prove that P and P' can be separated into the same 
number of tetrahedrons similar each to each and similarly 
placed. 

Proof. Let G and G ' be corresponding vertices. 

Divide all the faces of P and P', except those which include 
the angles G and G', into corresponding triangles by drawing 
corresponding diagonals. 

Pass a plane through G and each diagonal of the faces of P ; 
also pass a plane through G' and each corresponding diagonal 
of P'. 

Any two corresponding tetrahedrons G-ABC and G'-A'B'C' 
have the faces ABC, GAB, GBC similar respectively to the 
faces A'B'C', G'A'B', G'B'C. § 292 

AG _ AB_ _ AC_ __ BC_ GC_ 
6 A'G'~ A'B'~ A'C'~ B'C' G'C 1 ' * 

.-. the face GAC is similar to the face G'A'C § 289 



POLYHEDRONS 439 

They also have the corresponding trihedral . s equal. § 498 
.-. the tetrahedron G-ABC is similar to G'-A'B'C'. §710 

If G-ABC and G'-A'B'C are removed, the polyhedrons re- 
maining continue similar ; for the oew faces GAC and G'A'C 
have just been proved similar, and the modified faces .1 GF&nd 
A'G'F', GCHatod G'C'H', are similar (§ 292); also the iii.»di!i.Ml 
polyhedral AG and G', .1 and A' } C and C" remain equal each 
to each, since the corresponding parts taken from these angles 
are equal. 

The process of removing similar tetrahedrons can be carried 
on until the polyhedrons are separated into the same number 
of tetrahedrons similar each to each and similarly placed, q. e.d. 

718. Corollary 1. The corresponding edges of similar poly- 
hedrons are proportional. 

For the corresponding faces are similar. Therefore their correspond- 
ing sides are proportional (§ 282). 

719. Corollary 2. Any two corresponding lines in two 
similar polyhedrons hare the same rutin as any two corre- 
sponding edges. 

For these lines may be shown to be sides of similar polygons, and 
hence § 282 applies. 

720. Corollary 3. Two corresponding faces ,,f similar 
polyhedrons are proportional to the squares on ami two corre- 
sponding edges. 

For they are similar polyhedrons, and hence they are to each other 
as the squares on any two corresponding sides (§ 334). 

721. Corollarv 4. The entire surf ires of two similar poly- 
hedrons are proportional to the squares on any two correspond 
ing edges. 

For the corresponding faces are proportional to the squares i <n any fcw< i 
corresponding edges (§ 720), and hence their sum has the same proportion, 
by § 269. 



440 



APPENDIX TO SOLID GEOMETRY 



Proposition VI. Theorem 

722. The volumes of two similar tetrahedrons are to 
each other as the cubes on any two corresponding edges, 

v 




Given two similar tetrahedrons V-ABC and V'-A'B'C, with 
volumes v and v\ VB and V'B' being two corresponding edges. 



To prove that 



v 



VB" 



v' V'B 1 

Proof. Since the two polyhedrons are similar, Given 

.'. the corresponding polyhedral angles are equal, § 710 
and, in particular, the trihedral angles V and V' are equal. 

v VBX VC x VA 



v 



V'B' x VC x V'A' 

VB VC VA 
X T-rzr. X 



§714 



V'B' ' ' VC V'A' 
Furthermore, since the tetrahedrons are similar, 

VB VC VA 



V'B' VC VA 



'I A I 



Given 
§718 



VB 



Substituting —j— f for its equals, we have 
V B 



or 



v 
v' 

V 

v' 



VB VB VB 
X -7^-. X 



'W 



V'B' VB' 
VB Z 



inr 



Ax. 9 



Q. E. D. 



VB 



POLYHEDRONS 



441 



Proposition VII. Theorem 

723. The volumes of two similar polyhedrons are to 
each other as the cubes of ana tiro corresponding edges. 

L 




*K' 



l: \ 


s JI \ 




\'< A $ 


-fe] 


,— * \ 



B' 



Given two similar polyhedrons P and P f , with volumes v and v\ 
GB and G'B' being any two corresponding edges. 



To prove that v:v'= GB : G'B' . 

Proof. Separate P and P' into tetrahedrons similar each to 
each and similarly placed (§ 717), denoting their respective 
volumes by v v r , y gJ 

Then since 

o o 

§722 



•, v 1} r 2 , v 8} 



v : /■[ 



GB : G'B 



I r>l- 



I- 



vl = GB : G'B' , and so on. 



•'. >\+>- 2 +i\+---:<i+d + vl + ---=Glf: G'B'\ §269 



But 



'•i + r 2 + v s H = r, and v[ -f ''J -f »i + 



v 



.-. /• : y' = G'£' : G"YJ' , by Ax. 9. q.e.d. 

724. Prismatoid. A polyhedron having for bases two polygons 
in parallel planes, and for lateral faces triangles or trapezoids 
with one side common with one base, and the opposite vertex 
or side common with the other base, is called a prismatoid. 

The altitude is the distance between the plants of tin- bases. The mid- 
section is the section made by a plane parallel to the bases and bisecting 
the altitude. 



442 



APPENDIX TO SOLID GEOMETRY 



Proposition VIII. Theorem 

725. The volume of a prismatoid is equal to the prod- 
uct of one sixth of its altitude into the sum of its bases 
and four times its mid-section. 




Given a prismatoid of volume i>, bases b and &', mid-section m, 
and altitude a. 

To prove that v = J a (b -f- V + 4 rri). 

Proof. If any lateral face is a trapezoid, divide it-into two 
triangles by a diagonal. 

Take any point P in the mid-section and join P to the 
vertices of the polyhedron and of the mid-section. 

Separate the prismatoid into pyramids which have their 
vertices at P, and for their respective bases the lower base 
b, the upper base b\ and the lateral faces of the prismatoid. 

The pyramid P-XAB, which we may call a lateral pyra- 
mid, is composed of the three pyramids P-XQR, P-QBR, and 
P-QAB. 

Now P-XQR may be regarded as having vertex X and base 
PQR, and P-QBR as having vertex B and base PQR. 

Hence the volume of P-XQR is equal to i a ■ PQR, 
and the volume of P-QBR is equal to £ a • PQR. § 559 



POLYHEDRONS 443 

The pyramids P-QAB and P-QBR have the same vertex P. 
The base QAB is twice the base QBR (§ 327), since the A QAB 
has its base AB twice the base QR of the A QBR (§ 13G), and 
these triangles have the same altitude (§ 724). 

Hence the pyramid P-QA B is equivalent to twice the pyramid 

p-qbr. i :><;:; 



Hence the volume of P-QAB is equal to § a • PQR. 

Therefore the volume of P-XAB, which is composed of 
P-XQR, P-QBR, and P-QAB, is equal to f. a PQR. 

In like manner, the volume of each lateral pyramid is equal 
to | a x the area of that part of the mid-section which is 
included within it ; and therefore the total volume of all these 
lateral pyramids is equal to f am. 

The volume of the pyramid with base b is ^ ab, 
and the volume of the pyramid with base V is \ ab'. § 559 

Therefore v = \ a (Jj -f- V A- 4 in). q.e. d. 

EXERCISE 116 

Deduce from the formula for the volume of a prinmatoi > I . 
r = l a (b -f- b' + 4 ni), the following formulas : 

1. Cube, v = a 3 . 3. Pyramid, v = ^ ha. 

2. Prism, v = ba. 4. Parallelepiped, r = ba. 

5. Frustum of a pyramid, v = £ a-(fi + &' -f- v AA'). 

6. A prismatoid lias an upper base 3 sq. in., a lower base 
7 sq. in., an altitude 3 in., and a mid-section 4 sq. in. What 
is the volume ? 

7. A wedge has for its base a rectangle / in. long and w in. 
wide. The cutting edge is e in. long, and is parallel to the base. 
The distance from e to the base is d in. Deduce a f< irmula for the 
volume of the wedge. Apply this formula to the case in which 
I = 6. w = 1, e = 5, d = 3. 



444 



APPENDIX TO SOLID GEOMETEY 



Proposition IX. Theorem 

* 726. The volume of a spherical segment is equal to the 
product of one half the sum of its bases by its altitude, 
increased bij the volume of a sphere having that altitude 
for its diameter. 





Given a spherical segment of volume v, generated by the revo- 
lution of ABQP about MN as an axis, r being the radius of the 
sphere, AP being represented by r XJ BQ by r 2 , and PQ by a. 

To prove that v = 1 a (yrr* -f- 7rr 2 2 ) + 1 ira z . 

Proof. We shall first find the volume of the spherical seg- 
ment with one base, generated by AMP. 

Area of zone AM— 2 irr ■ PM. § 691 

. ' . volume of sector generated by OA M=±rx2irr- PM. §708 

But the cone generated by OA P = \ irr} (r—PM ). § 611 

.'. volume AMP = * r X.2 irr • PM- — i Trr*(r—PM). Ax. 2 

But r* = PM x NP = PM (2 r - PM). § 297 

.'. volume AMP = i r x 2 irr • PM 



= tt-PM ( 



i 7T • PM (2 r — PAQ (r — PM ) Ax. 9 

2 



In the same way, volume BMQ = it • Qilf (r — i QAf) 
.-. v = volume ^4ilfP — volume BMQ 



A 



= 7T • Pilf • r — 1 7T • 7W — 7T • (23/ • r + £ 7T • (2 AT 

= irr(PM 2 — QM 2 ) - | 7r(iW 8 - OM" 3 ). 



SPHERICAL SEGMENTS 445 

But PM — QM = a. Given 

r.v = 7rra(PM+QM)-$ira(PM 2 +PMQM+QM*). Ax. 9 
But a* = PM* — 2PM> QM+ QM 2 . Ax. 5 



... a * + 3PM> QM=PM +PM- QM+QM . Ax. 1 

.-. r = irra ( PM + QM) - } tt" (a a + 3 PM • QM). Ax. 9 
Furthermore (2 r - PM) PM = >- 2 , 

ami (2r-QM)QM=r|. §297 



'. 2 r • PAf -f- 2 r • QM - PM - QM = rf + rl Ax. 1 
,\r>PM+rQM= T H 77 Axs. 1, 4 



.. v== ,ra( — — + g --PM.QM 

= tt« ^-±^ + |* + PM- QM - f - PM- QM 

= |ffi (7T7'! 2 -f 7ivf) + I 7Ta* Q- E. D. 

EXERCISE 117 

Find the volumes of sj)herical segments having bases b and 
b', and altitudes a, as follows : 

1. b = ±,b' = 5,a = l. 4. b = 6, b'= 8, a = If 

2. ft = 4, b' = 6, a = 1 -\. 5. b = 8, b'= 12, a = 2. 

3. b = 5, 6' = 7, a = 2J. 6. 5 = 12, &'= 15, a = 3£. 

7. b = 27 s<|. in., // = 32 sq. in., a = 2.33 in. 

Find the volumes of spherical segments ha ring radii of ba±> % 
r and r 2 , and altitudes a, as follows: 

8. r x = 3, r a = 4, a = 2. 11. r t = 5, r 2 = 3, a = l£. 

9. r, = 4, /•., = 7, a = 3. 12. r 1= G, r 2 = 5, a = 1 j . 
10. r, = 8, r a = 5, a = 4£. 13. r 1= 9, r = 10, a = 2f . 

14. /-j = 9 in., r 2 = 7 in., « = 4.75 in. 



446 APPENDIX TO SOLID GEOMETRY 

EXERCISE 118 
Examination Questions 

1. A pyramid 6 ft. high is cut by a plane parallel to the 
base, the area of the section being | that of the base. How far 
from the vertex is the cutting plane ? 

2. Find the area of a spherical triangle whose angles are 
100°, 120°, and 140°, the diameter of the sphere being 16 in. 

3. Two angles of a spherical triangle are 80° and 120°. 
Find the limits of the third angle, and prove that the greatest 
possible area of the triangle is four times the least possible 
area, the sphere on which it is drawn being given. 

4. An irregular portion, less than half, of a material sphere 
is given. Show how the radius can be found, compasses and 
ruler being allowed. 

5. Find the volume of a cone of revolution, the area of 
the total surface of which is 200 it sq. ft., and the altitude of 
which is 16 ft. 

6. The volumes of two similar polyhedrons are 64 cu. ft. 
and 216 cu. ft. respectively. If the area of the surface of the 
first polyhedron is 112 sq. ft., find the area of the surface of 
the second polyhedron. 

7. A solid sphere of metal of radius 12 in. is recast into a 
hollow sphere. If the cavity is spherical, of the same radius 
as the original sphere, find the thickness of the shell. 

8. The stone spire of a church is a regular pyramid 50 ft. 
high on a hexagonal base each side of which is 10 ft. There 
is a hollow part which is also a regular pyramid 45 ft. high, on 
a hexagonal base of which each side is 9 ft. Find the number 
of cubic feet of stone in the spire. 

9. The volumes of a hemisphere, right circular cone, and 
right circular cylinder are equal. Their bases are also equal, 
each being a circle of radius 10 in. Find the altitude of each. 



EXERCISES 447 

10. A sphere of radius 5 ft. and a right circular cone also of 
radius 5ft. stand on a plane If the heighl of the cone is 
equal to a diameter of the sphere, find the position of the plane 
that cuts the two solids in equal circular sections. 

11. The vert ices of one regular tel rahedron are at the centers 
of the faces of another regular tetrahedron. Find the ratio of 
the volumes. 

12. Find the area of a spherical triangle, if the perimeter of 

its polar triangle is L".)7° and the radius of the sphere is 10 
centimeters. 

13. The radii of two spheres are 13 in. and 15 in. respec- 
tively, and the distance between the centers is 14 in. Find the 
volume of the solid common to both spheres, — a spherical lens. 

14. The radius of the base of a right circular cylinder is r 
and the altitude of the cylinder is a. Find the radius and the 
volume of a sphere whose surface is equivalent to the lateral 
surface of the cylinder. 

15. If the polyhedral angle at the vertex of a triangular 
pyramid is trirectangular, and the areas of the lateral faces 
are a, 5, and c respectively, and the area of the base is '/. 
then a 2 + V 1 + c 2 = d 2 . 

16. If the earth is a sphere with a diameter of 8000 mi., 
find the area of the zone bounded by the parallels 30° north 
latitude and 30° south latitude Show that this zone and the 
planes of the circles include {£ of the volume of the earth. 

17. The altitude of a cone of revolution is 12 centimeters 
and the radius of it-- base is 5 centimeters. Compute the radius 
of the sector of paper which, when rolled up, will just cover 
the convex surface of the cone, and compute the size of the 
central angle of this sector in degrees, minutes, and seconds. 

18. The volume of any regular pyramid is equal to one 
third of its lateral area multiplied by the perpendicular dis- 
tance from the center of its base to any lateral face. 



448 APPENDIX TO SOLID GEOMETRY 

19. If the area of a zone of one base is n times the area of 
the circle which forms its base, the altitude of the zone is 

: (n — 1) times the diameter of the sphere. Discuss the special 

case when n = 1. 

20. If the four sides of a spherical quadrilateral are equal, 
its diagonals are perpendicular to each other. 

21. Find the volume of a pyramid whose base contains 30 
square centimeters if one lateral edge is 5 centimeters and the 
angle formed by this edge and the plane of the base is 45°. 

22. On the base of a right circular cone a hemisphere is 
constructed outside the cone. The surface of the hemisphere 
equals the surface of the cone. If r is the radius of the hemi- 
sphere, find the slant height of the cone, the inclination of the 
slant height to the base, and the volume of the entire solid. 

23. Find the total, surface and the volume of a regular tetra- 
hedron whose edge equals 8 centimeters. 

24. If a spherical quadrilateral is inscribed in a small circle, 
the sum of two opposite angles is equal to the sum of the 
other two angles. 

25. By what number must the dimensions of a cylinder of 
revolution be multiplied to obtain a similar cylinder of revo- 
lution with surface n times that of the first ? with volume n 
times that of the first ? 

26. A pyramid is cut by a plane parallel to the base midway 
between the vertex and the plane of the base. Compare the 
volumes of the entire pyramid and the pyramid cut off. 

27. The height of a regular hexagonal pyramid is 36 ft. and 
one side of the base is 6 ft. What are the dimensions of a 
similar pyramid whose volume is J^ that of the first ? 

28. One of the lateral edges of a pyramid is 4 meters. How 
far from the vertex will this edge be cut by a plane parallel to 
the base, which divides the pyramid into two equivalent parts ? 



RECREATIONS 



449 



727. Recreations of Geometry. The following simple puzzles 
and recreations of geometry may serve the double purpose of 
adding interest to the study of the subject and of leading the 
student to exercise greater care in his demonstrations. They 
have long been used for this purpose and are among the best 
known puzzles of geometry. 




EXERCISE 119 

1. To prove that every triangle is isosceles. 

Let ABC be a A that is not isosceles. 

Take CP the bisector of ZACB, and ZP the ± bisector of AB. 

These lines must meet, as at P, for otherwise 
they would be II, which would require CP to be _L 
to AB, and this could happen only if A ABC were 
isosceles, which is not the case by hypothesis. 

From P draw PX _L to BC and PY ± to CA, and 
draw PA and PB. 

Then since ZP is the J_ bisector of AB, .-. PA = PB. 

And since CP is the bisector of Z ACB, .-. PA' = PY. 

.-. the rt. 4 PBX and PA Y are congruent, and BX = AY. 

But the rt. & PXC and PYC are also congruent, and .-. XC =YC. 

Adding, we have BX + AC = A Y + YC, or BC = A C. 

.-. A ABC is isosceles, even though constructed as not isosceles. 

2. To prove that part of an angle equals the whole angle. 
Take a square ABCD, and draw MWP, the _L bisector of CD. Then 

A/ATP is also the _L bisector of AB. 

From B draw any line BX equal to AB. 

Draw BX and bisect it by the A XP. 

Since DA" intersects CD, Js to these lines can- 
not be parallel, but must meet as at P. 

Draw PA, PD, PC, PX. and PB. 

Since _VP is the A bisector of CD, PD = PC. 
Similarly PA = PB, and PD = PX. 

.-. PX = PD = PC. 

But BX = BC by construction, and PB is common to & PBX and PBC. 

.-. A PBX is congruent to A PB( '. and Z XBP = Z CBP. 

.-. the whole Z XBP equals its i art, the Z.CBP. 




-iX 



450 



APPENDIX TO SOLID GEOMETRY 



3. To prove that part of an angle equals the whole angle. 

Take a right triangle ABC and con- 
struct upon the hypotenuse BC an equi- 
lateral triangle BCD, as shown. 

On CD lay off CP equal to CA. 

Through X, the mid-point of AB, 
draw PX to meet CB produced at Q. 
Draw QA. 

Draw the _L bisectors of QA and 
QP, as YO and ZO. These must meet 
at some point because they are ± to 
two intersecting lines. 

Draw OQ, OA, OP, and OC. 

Since is on the ± bisector of QA, .-. OQ = OA. 

Similarly OQ= OP, and .-. OA = OP. 

But CA = CP, by construction, and CO = CO. 

.-. A A OC is congruent to A POC, and ZACO=Z PCO. 

4. To prove that part of a line equals the whole line. 

Take a triangle A BC and draw CP ± to AB. c 

From C draw CX, making Z ACX = Z B. 
Then A ABC and ACX are similar. 




.-. AABC:AACX= BC -.CX\ 
Furthermore A ABC: A ACX = AB : AX. 




^2 



or 

But 
and 



BC'-.CX =AB:AX, 
BC 2 :AB= CX 2 :AX. 



BC= AC + AB' - 2 AB ■ AP, 
CX 2 = AC 2 + AX 2 - 2 AX ■ AP. 



AC + AB--2AB- AP AC + AX' -2AX- AP 



AB 



or 



AC 



;2 



or 



AB ' 


/1X) — 


£ ^XJT 


•'• 


ZC 2 


AX 


Zcr 


J -^B 


■ AX 



AC 
AX 
AC 2 
AX 



AX 



+ AX-2AP. 



- AB. 



AC - AB • AX 



AB 



AX 



AB = AX. 



RECREATIONS 



451 



5. To show geometrically that 1=0. 

Take a square that is 8 units on a side, and cut it into three parts, 
A, B, C, as shown in the right-hand figure. Fit 
these parts together as in the left-hand figure. 

Now the square is 8 units on aside, and therefore 

contains 8 x 8, or 04, small squares, while the rec- 
tangle is 13 units long and 5 units high, and there- 
fore contains 



i i i 

H-+-+— ' — i- 

H + - + - 

I I I 




5 x 13, or 65, 




-4—4--+— }-- H — 

-4— +-^--4—1— 4— 
I C| 

— I — I — h— \—A— A — \- 
i i i I i I i 
-H — 4- — | — f~J — I — I — 

I < I I I I 



small squares. 
But the two 
figures are each made up of A + B+ C 
i Ax.ll ), and therefore areequal(Ax.8). 
65 = 64, and by subtracting 64 we have 1 = (Ax. 2). 



6. To prove that any point on a line bisects it. 

Take any point P on AB. 

On AB construct an isosceles A ABC, having 
AC = BC; and draw PC. 

Then in AAPC and PBC, we have 

Z A = Z B, 

AC= BC, 

and PC = PC. 



Iiavin° r 






>\ 








' ! \ 






» 
» 
t 


\ 






1 
t 


\ \ 


§74 




/ 
/ 
/ 
/ 
t 


\ \ 
i \ 
i \ 

i \ 


Const. 




t 

t 
/ 


\ \ 




A' 




! 


Iden. 




./' 



/; 



Three independent parts (that is, not merely the three angles) of one 
triangle are respectively equal to three parts of the other, and the tri- 
angles are congruent ; therefore AP = BP (§ 07). 

7. To prove that it is possible to let fall two perpendiculars 
to a line from an external point. 

Take two intersecting (D with centers and C . 

Let one point of intersection be P. and draw the diameters PA and PD. 

Draw AB cutting the circumferences at B 
and C. Then draw PB and PC. 

Since Z. PC A is inscribed in a semicircle, 
it is a right amrle. In the same way, since 
ZIjP>P is inscribed in a semicircle, it also is '' 
a right angle. 

.-. PJJand PC are both L to AD. 




452 



APPENDIX TO SOLID GEOMETRY 




8. To prove that if two opposite sides of a quadrilateral are 
equal the figure is an isosceles trapezoid. 

Given the quadrilateral ABCD, with BC = DA. 
To prove that AB is || to DC. 

Draw MO and NO, the ± bisectors of AB and 
CD, to meet at 0. 

If AB and DC are parallel, the proposition is already proved. 

If AB and DC are not parallel, then MO and NO will meet at 0, either 
inside or outside the figure. Let be supposed to be inside the figure. 

Draw OA, OB, 0(7, OD. 
Then since OM is the JL bisector of AB, .-. OA = OB. 
Similarly OD = OC. 

But DA is given equal to BC. 
.-. A A OD is congruent to ABOC, 
ZD0A = ZB0C. 

* 

rt. A OCN and ODiV are congruent, 
ZN0D = ZC0N. 

rt. A AM and Z>3/0 are consruent, 
Z.10AT = ZiY0£. 
.-. Z NOD + Z DCvl + Z A OM = Z CON + Z BOC + Z MOB, 
or Z NOM = Z MON = a st. Z. 

Therefore the line M ON is a straight line, and hence AB is II to DC. 

If the point is outside the quadrilateral, as 
in the second figure, the proof is substantially the 
same. 

Tor it can be easily shown that 

Z DON - Z DOA - Z A OM 

= Z NOC - Z BOC - Z MOB, 
which is possible only if 

ZD0N=ZD0M, 
or if ON lies along OM. 

But that the proposition is not true is evident from the 
third figure, in which BC = DA, but AB is not II to DC. 



and 

Also, 
and 

Similarly 
and 





HISTORY OF GEOMETRY 453 

728. History of Geometry. The geometry of very ancient 
peoples was largely the mensuration of simple arras and 
volumes such as is taught to children in elementary arithmetic 
to-day. They learned how to tind the area of a rectangle, 

and in the oldest mathematical records that we have there is 
some discussion of triangles and of the volumes of solids. 

The earliest documents that we have, relating to geometry, 
come to us from Babylon and Egypt. Those from Babylon 
were written about 2000 B.C. on small clay tablets, some of 
them about the size of the hand, these tablets afterwards 
having been baked in the sun. They show that the Baby- 
lonians of that period knew something of land measures, and 
perhaps had advanced far enough to compute the area of a 
trapezoid. For the mensuration of the circle they later used, 
as did the early Hebrews, the value it = 3. 

The first definite knowledge that we have of Egyptian math- 
ematics comes to us from a manuscript copied on papyrus, a 
kind of paper used about the Mediterranean in early times. 
This copy was made by one Aah-mesu (The Moon-born), com- 
monly called Ahmes, who. probably nourished about 1700 b.c. 
The original from which he copied, written about 2300 b.c, 
has been lost, but the papyrus of Ahmes, written nearly four 
thousand years ago, is still preserved and is now in the British 
Museum. In this manuscript, which is devoted chiefly to frac- 
tions and to a crude algebra, is found some work on mensu- 
ration. Among the curious rules are the incorrect ones that 
the area of an isosceles triangle equals half the product of 
the base and one of the equal sides; and that the area of a 
trapezoid having bases h, b\ and nonparallel sides each equal 
to ". is I a(b-\-b'). One noteworthy advance appears however. 
Ahmes gives a rule for finding the area of a circle, substan- 
tially as follows: Multiply the square on the radius by (\f') 2 , 
which is equivalent to taking for it the value 3.1605. Long 
before the time of Ahmes, however, Egypt had a good working 



454 APPENDIX TO SOLID GEOMETEY 

knowledge of practical geometry, as witness the building of 
the pyramids, the laying out of temples, and the digging of 
irrigation canals. 

Erom Egypt and possibly from Babylon geometry passed to 
the shores of Asia Minor and Greece. The scientific study of 
the subject begins with Thales, one of the Seven Wise Men 
of the Grecian civilization. Born at Miletus about 640 B.C., 
he died there in 548 B.C. He spent his early manhood as a 
merchant, accumulating the wealth that enabled him to spend 
his later years in study. He visited Egypt and is said to have 
learned such elements of geometry as were known there. He 
founded a school of mathematics and philosophy at Miletus, 
known as the Ionic School. How elementary the knowledge 
of geometry then was, may be understood from the fact that 
tradition attributes only about four propositions to Thales, 
substantially those given in §§ 60, 72, 74, and 215 of this book. 

The greatest pupil of Thales, and one of the most remark- 
able men of antiquity, Avas Pythagoras. Born probably on the 
island of Samos, just off the coast of Asia Minor, about the 
year 580 B.C., Pythagoras set forth as a young man to travel. 
He went to Miletus and studied under Thales, probably spent 
several years in study in Egypt, very likely went to Babylon, 
and possibly went even to India, since tradition asserts this 
and the nature of his work in mathematics confirms it. In 
later life he went to southern Italy, and there, at Crotona, in 
the southeastern part of the peninsula, he founded a school 
and established a secret society to propagate his doctrines. 
In geometry he is said to have been the first to demonstrate 
the proposition that the square on the hypotenuse of a right 
triangle is equivalent to the sum of the squares on the other 
two sides (§ 337). The proposition was known before his time, 
at any rate for special cases, but he seems to have been the 
first to prove it. To him or to his school seems also to have 
been due the construction of the regular pentagon (§§ 397, 398) 



HISTORY OF GEOMETRY 455 

ar.d of the five regular polyhedrons. The construction of the 
regular pentagon requires the dividing of a line in extreme 
and mean ratio (§ 311), and this problem is commonly assigned 
to the Pythagoreans, although it played an important part in 
Plato's school. Pythagoras is also said to have known thai six 
equilateral triangles, three regular hexagons, or four squari . 
«-an be placed about a point so as jusl to till the 360°, bul thai 
no other regular polygons can be so placed. To his school is 
also due the proof thai the sum of the angles of a triangle 
equals two right angles (§ 107), and the construction of at 
least one star-polygon, the star-pentagon, which became the 
badge of his fraternity. 

For two centuries after Pythagoras geometry passed through 
a period of discovery of propositions. The state of the science 
may be seen from the fact that (Enopides of Chios, who 
flourished about 465 B.C., showed how to let fall a perpendicu- 
lar to a line (§ 227), and how to construct an angle equal to a 
given angle (§ 232). A few years later, about 440 B.C., Hippoc- 
rates of Chios wrote the first Greek textbook on mathematics. 
He knew that the areas of circles are proportional to the squares 
on their radii, but was ignorant of the fact that equal central 
angles or equal inscribed angles intercept equal arcs. 

About 430 B.C. Antiphon and Bryson, two Greek teachers. 
worked on the mensuration of the circle. The former attempted 
to find the area by doubling the number of sides of a regular 
inscribed polygon, and the latter by doing the same for both in- 
scribed and circumscribed polygons. They thus substantially 
exhausted the area between the circle and the polygon, and 
hence this method was known as the Method of Exhaustions. 

During this period the great philosophic school of Plato 

129-348 b.c.) flourished at Athens, and to this school is due 
the first systematic attempt to create exact definitions, axioms, 
and postulates, and to distinguish between elementary and 
higher geometry. At this time elementary geometry became 



456 APPENDIX TO SOLID GEOMETRY 

limited to the use of the compasses and the unmarked straight- 
edge, which took from this domain the possibility of con- 
structing a square equivalent to a given circle ("squaring the 
circle"), of trisecting any given angle, and of constructing 
a cube with twice the volume of a given cube ("duplicating 
the cube "), these being the three most famous problems of 
antiquity. Plato and his school were interested in the so-called 
Pythagorean numbers, numbers that represent the three sides 
of a right triangle. Pythagoras had already given a rule to 
the effect that J (m 2 + l) 2 = m 2 + J (m 2 - l) 2 . The school of 
Plato found that [(|m) 2 + l] 2 =m 2 i[(|m) 2 -l] 2 . By giving 
various values to m, different numbers will be found such that 
the sum of the squares of two of them is equal to the square of 
the third. 

The first great textbook on geometry, and the most famous 
one that has ever appeared, was written by Euclid, who taught 
mathematics in the great university at Alexandria, Egypt, 
about 300 b.c. Alexandria was then practically a Greek city, 
having been named in honor of Alexander the Great, and 
being ruled by the Greeks. 

Euclid's work is known as the M Elements," and, as was the case 
with all ancient works, the leading divisions were called books, 
as is seen in the Bible and in such Latin writers as Csesar 
and Vergil. This is why we speak of the various books of 
geometry to-day. In this work Euclid placed all the leading 
propositions of plane geometry as then known, and arranged 
them in a logical order. Most subsequent geometries of any im- 
portance since his time have been based upon Euclid, improving 
the sequence, symbols, and wording as occasion demanded. 

Euclid did not give much solid geometry because not much 
was known then. It was to Archimedes (287-212 b.c), a 
famous mathematician of Syracuse, on the island of Sicily, 
that some of the most important propositions of solid geometry 
are due, particularly those relating to the sphere and cylinder. 



HISTORY OF GEOMETRY 457 

He also showed how to find the approximate value of w by a 
method similar to the one we teach to-day (§404), proving 
that the real value lies between 3} and 3i f >. Tradition says 
that the sphere and cylinder were engraved upon his tomb. 
The Greeks contributed little more to elementary geometry, 
although Apollonius of Perga, who taught at Alexandria be- 
tween 250 and 200 n.c. wrote extensively on conic sect ions; and 
Heron of Alexandria, about the beginning of the Christian era, 
showed that the area of a triangle whose sides are »i. /,. ,-. equals 
V* (*■ — a) (n — b) (s — c), where s = ±(a+b + c) (see p. 211). 

The East did little for geometry, although contributing 
considerably to algebra. The first great Hindu writer was 
Aryabhatta, who was born in 47G a.d. He gave the very 
close approximation for 77-, expressed in modern notation as 
3.1416. The Arabs, about the time of the Arabian Nights Tales 
(800 a.d.), did much for mathematics, translating the Greek 
authors into their own language and also bringing learning 
from India. Indeed, it is to them that modern Europe owes 
its first knowledge of Euclid. They contributed nothing of 
importance to geometry, however. 

Euclid was translated from the Arabic into Latin in the 
twelfth century, Greek manuscripts not being then at hand, or 
b.-ing neglected because of ignorance of the language. The 
leading translators were Athelhard of Bath (1120), an English 
monk who had learned Arabic in Spain or in Egypt; Gerhard 
of Cremona, an Italian monk; and Johannes Campanus, chap- 
lain to Pope Urban IV. 

In the Middle Ages in Europe nothing worthy of note was 
added to the geometry of the Greeks. The first edition of 
Euclid was printed in Latin in 1482, the first one in English 
appearing in 1570. Our symbols are modern, + and — first 
appearing in a German work in 1489; = in Recorded "Whet- 

>ne of Witte" in 1557; > and < in the works of Harriot 
(1560-1621); and x in a publication by Oughtred (1574-1660). 



458 APPENDIX TO SOLID GEOMETRY 

729. Areas of Solid Figures. The following are the more 
important areas of solid figures : 

Prism, l=ep (§512). 

Regular pyramid, l—'Lsp (§ 553). 

Frustum of regular pyramid, / — \ (jp +^')s (§ 554). 

Cylinder of revolution, I = ac '= 2 irra (§ 588). 

Cone of revolution, / = \sc = irrs (§ 609). 

Frustum of cone of revolution, / = i (c -f- c') s (§ 615). 

Sphere, s = 4 tt/- 2 (§ 689). 

Zone, 5 = 2 7rra (§ 691). 

Z,4 2 Z,4 

360 90 



Lune, s = — • 4 tt/- 2 = — — • 7T7- 2 (§ 694). 



730. Volumes. The following are the more important volumes : 

Rectangular parallelepiped, liva (§ 534). 

Prism or cylinder, ba (§§ 539, 589). 

Pyramid or cone, \ba (§§ 561, 611). 

Frustum of pyramid or cone, \ a (p -\-b' -\- vW) (§§ 565, 617). 

Right-circular cylinder, irra (§ 590). 

Cone of revolution, \ iri^a (§ 612). 

Frustum of cone of revolution, \ 7r«(V 2 + r' 2 -f- rr 1 ) (§ 618). 

Prismatoid, J a (6 + &' + 4 m) (§ 725). 

Sphere, f vrr 3 = J ird 3 (§ 706). 

Spherical pyramid, \ br. 

Spherical sector, i zr (§ 708). 

Spherical segment, \ a (irrl -j- 7rr|) -+- £ 7ra 3 (§ 726). 



INDEX 



PAGl PA( 

Altitude of cone 362 Classes of polyhedral angles . 308 

of cylinder ; >-">:; Concave polyhedral angle . 3 

Cone 362 

altitude of 362 

axis Of :;,;; ' 

base of 362 

circular 363 

circumscribed 366 

element of '■'>'''- 

frustum of 367 

inscribed 366 

lateral surface of . . . . 302 

oblique : '> lV ■'• 

of revolution 363 

right \ . . 363 

slant height of cone of revo- 
lution '•'>'''■'> 

vertex of 362 

(Ones 363 

. 367 



of frustum of cone . . . 367 
of frustum of pyramid . . 338 

of prism 317 

of prismatoid 441 

of pyramid 337 

of spherical segment . .421 
of zone 410 

Angle, dihedral 293 

of lune 410 

polyhedral 308 

spherical 389 

tetrahedral 308 

trihedral 308 

Axis of circular cone .... 363 

Base of cone 302 

. 421 
. 421 

. 353 
. 367 
. 338 



of spherical pyramid 
of spherical sector . . 
bs of cylinder .... 
of frustum of cone . . 
of frustum of pyramid . 

of prism 317 

of spherical segment . . .421 

of zone 41 1 > 

prisms classified as to . . 318 
pyramids classified as to . 337 

( Circles, describing, on sphere ■ - 
( !ircumscribed prism .... 356 

pyramid : 

sphere 386 



and frustums as limits . 

similar 370 

Congruent solids 322 

spherical polygons . . . 392 

( tonic surface 362 

directrix <>t* 362 

dement of '■'>''>- 

neratrix of 362 

lower nappe of 362 

upper nappe of 362 

section ;;,;; > 

( tonstruction of tangent plan 

to cones 

to cylinders 






400 



INDEX 



PAGE 

Convex polyhedral angle . . 308 

polyhedron 317 

spherical polygon .... 302 

Cube 322 

Cylinder 353 

altitude of 353 

as a limit 357 

bases of 353 

circular 354 

circumscribed 356 

inscribed 356 

lateral surface of ... . 353 

oblique 353 

of revolution 354 

right 353 

right section of 357 

section of 353 

tangent plane to .... 356 

Cylinders, similar 350 

Cylindric surface 353 

directrix of 353 

element of 353 

generatrix of 353 

Describing circles on sphere . 385 

Diagonal, spherical 302 

Diameter of sphere 381 

Dihedral angle 203 

acute 203 

edge of 203 

faces of 203 

obtuse 203 

plane angle of 204 

reflex 203 

right 203 

size of 203 

straight 203 

Dihedral angles, adjacent . . 203 

complementary 203 

conjugate 203 



PAGE 

Dihedral angles, relation to 

plane angles 204 

supplementary 203 

vertical 203 

Dimensions 320 

Distance from a point to a 

plane 270 

polar 384 

spherical -. . 383 

Dodecahedron, regular . . . 351 

Edge of dihedral angle . . . 203. 

Edges of polyhedral angle . . 308 

polyhedron 317 

Element of conic surface . . 362 

cylindric surface .... 353 

Ellipse 363 

Equal polyhedral angles . . 308 

Equivalent solids 322 

Euler's theorem 435 

Excess, spherical, of poly- 
gon 417 

of triangle 413 

Face angle of polyhedral angle 308 

Faces of dihedral angle . . . 203 

polyhedral angle .... 308 

polyhedron 317 

polyhedrons classified as to 350 

Foot of line 273 

Frustum of cone 367 

altitude of 367 

bases of 367 

lateral area of 367 

slant height of 367 

Frustum of pyramid .... 338 

altitude of 338 

lateral area of 338 

lateral faces of 338 

slant height of 338 



INDEX 



461 



PAGE 

Generation of spherical surface 381 
zone 410 

Hemisphere 381 

Bexahedron, regular .... 3">l 
Hyperbola 363 

[cosahedron, regular .... 351 
Inclination of a line .... 303 
Inscribed polyhedron . . . . 386 

prism 356 

pyramid 366 

sphere 386 

Lateral ana of frustum . . . 338 

prism 317 

pyramid 337 

edges of prism 317 

pyramid 337 

faces of frustum .... 338 

prism 317 

pyramids 337 

surface of cone 3G2 

cylinder 353 

frustum of cone .... 367 
Limit, cylinder as a ... . 357 
Limits, cones and frustums as 367 
Line and plane parallel . . . 282 

foot of 273 

inclination of 303 

oblique 277 

projection of 302 

Lune 410 

angle of 410 

Mid-section of prismatoid . . 441 

Nappes of conic surface . . 362 

( iblique line ^77 

and right cylinders . . . 353 
Octahedron, regular .... 351 



PAGE 

Parabola 363 

Parallel line and plane . . . 2S2 
planes 285 

Parallelepiped 322 

rectangular 322 

right 322 

Perpendicular planes .... 293 
to a plane -7"> 

Plane 273 

angle of dihedral angle. . 2 ( .»1 

determining 273 

perpendicular to ... . 27"> 

Planes, intersection of . . . 273 

parallel 285 

perpendicular 293 

postulate of 274 

Point, projection of a ... 302 

Polar distance 384 

triangle . 394 

Poles of a circle 383 

Polygon, angles of 392 

excess of 417 

sides of 392 

spherical 392 

vertices of 392 

Polyhedral angle 308 

concave 308 

convex 308 

edges of 308 

face angles of 308 

faces of 308 

parts of 308 

size of 308 

vertex of ....... 308 

Polyhedral angles 308 

classes of 308 

equal 308 

symmetric 311 

Polyhedron 317 

convex 317 



462 



INDEX 



TAGE 

Polyhedron, edges of . . . .317 

faces of 317 

regular 350 

section of 317 

vertices of 317 

Polyhedrons classified as to 

faces 350 

.similar 431 

Postulate of planes .... 274 

Prism 317 

altitude of 317 

bases of 317 

circumscribed 356 

inscribed 356 

lateral area of 317 

lateral edges of 317 

lateral faces of 317 

oblique 318 

right 318 

right section of 318 

truncated 318 

Prismatoid 441 

altitude of 441 

mid-section of 441 

Prisms classified as to bases . 318 

Projection of a line .... 302 
of a point 302 

Pyramid 337 

altitude of 337 

base of . 337 

circumscribed 366 

frustum of 338 

inscribed 366 

lateral area of 337 

lateral edges of 337 

lateral faces of 337 

quadrangular 337 

regular 337 

right 337 

slant height of regular . .337 



PAGE 

Pyramid, spherical 421 

triangular 337 

vertex of 337 

Pyramids classified as to bases 337 
properties of regular . . . 338 

Quadrant 384 

Radius of sphere 381 

Regular polyhedron .... 350 

pyramid 337 

Relation of dihedral angles to 

plane angles 294 

polygons to polyhedral 

angles 392 

Revolution of cone 363 

Right and oblique cones . . . 363 

cylinders 353 

Right prism ....... 318 

Right section of cylinder . . 357 

of prism 318 

Section of a cone 363 

cylinder ........ 353 

polyhedron 317 

right, of a cylinder . . . 357 

prism 318 

Sector, spherical 421 

Segment, spherical 421 

Similar cones 370 

cylinders 359 

polyhedrons 431 

Size of polyhedral angle . . 308 
Slant height of cone of revolu- 
tion 363 

frustum of cone of revolu- 
tion 367 

frustum of pyramid . . . 338 

regular pyramid .... 337 

Solids, congruent 322 



INDEX 



4»;:; 



PAGE 

Solids, equivalent 322 

Sphere 381 

center of 381 

circumscribed 380 

describing circles on . . . 385 

diameter of 381 

great circle of 383 

inscribed 380 

poles of a circle 383 

radius of 381 

small circle of 383 

Spheres, tangent 385 

Spherical angle 389 

distance 383 

excess of a triangle . . . 413 

polygon 417 

convex 392 

diagonal of 302 

polygons, congruent . . . 302 

pyramid 421 

base of 421 

vertex of 421 

sector 421 

base of 421 

segment 421 

altitude of 421 

bases of 421 

of one base 421 

surface, generation of . . 381 

Spherical triangle 302 

diagonal of 302 

wedge 421 

Surface, conic 3<i2 

cylindric 353 

spherical 381 

Symmetric isosceles triangles . 399 
triangles, relation of . . . 399 
polyhedral angles .... 311 
spherical triangles . . . . 399 



PA < . 1 

Tangent plane to cone . . . 366 

to cylinder 356 

lines and planes .... 385 

spheres 385 

Tetrahedral angle 308 

Tetrahedron, regular .... 351 

Triangle, polar 394 

spherical 3'.>-2 

excess of A\:\ 

Triangles, birectangular . . 397 

classified as to right angles 397 

relation of symmetric . . 399 

symmetric isosceles . . . 399 

spherical 399 

trirectangular 397 

Trihedral angle 308 

Truncated prism 318 

Unit of volume 322 



Vertex of polyhedral angle . 308 

of spherical pyramid . . 421 

Vertices of a polyhedron . .317 

Volume 322 

unit of 322 

Wedge, spherical 42] 

Zone 410" 

altitude of 410 

bases of 410 

aeration of 41<> 

of one base 4 10 



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